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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

  • For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
  • For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

  • Add or Subtract the same value from both sides
  • Clear out any fractions by Multiplying every term by the bottom parts
  • Divide every term by the same nonzero value
  • Combine Like Terms
  • Expanding (the opposite of factoring) may also help
  • Recognizing a pattern, such as the difference of squares
  • Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

  • solve Quadratic Equations
  • solve Radical Equations
  • solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

  • Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
  • Show all the steps , so it can be checked later (by you or someone else)

Equation Solver

What do you want to calculate.

  • Solve for Variable
  • Practice Mode
  • Step-By-Step

Example (Click to try)

How to solve your equation, solving equations video lessons.

  • Solving Simple Equations

Need more problem types? Try MathPapa Algebra Calculator

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Problem solving with equations

Solving a task usually consists of finding the value of some quantity by logical reasoning and calculations. For example, find the speed, time, distance, mass of an object, or amount of something.

Such a task can be solved by using an equation. To do this, the desired value is denoted by a variable, then by logical reasoning the equation is composed and solved. After solving the equation, you check to see if the solution to the equation satisfies the conditions of the task.

Writing expressions containing the unknown

The solution of the task is accompanied by the composition of the equation to this task. At the initial stage of studying tasks it is desirable to learn to make up letter expressions describing this or that situation in life. This stage is not complicated and can be studied in the process of solving the task itself.

Consider a few situations that can be written down using a mathematical expression.

Task 1. The father is x years old. Mom is two years younger. The son is three times younger than the father. Write the age of each using expressions.

age of mother's father and son table 1

Task 2. Father is x years old, mother is 2 years younger than father. Son is 3 times younger than father, daughter is 3 times younger than mother. Write the age of each using expressions.

age of the father of the mother of the son and daughter table 2

Task 3. Father is x years old, mother is 3 years younger than father. The son is 3 times younger than the father, the daughter is 3 times younger than the mother. How old are each of them if the total age of the father, mother, son, and daughter is 92?

In this task, in addition to writing down the expressions, we need to calculate the age of each family member.

First, write down the age of each family member using expressions. We will take the age of the father as the variable x, and then use that variable to make the rest of the expressions:

age of the father of the mother of the son and daughter table 3

Now let's determine the age of each family member. To do this, we need to make and solve an equation. We have all the components of the equation ready. The only thing left to do is to put them together.

The total age of 92 was obtained by adding the ages of the father, mother, son and daughter:

father + mother + son + daughter = 92

For each age, we made up a mathematical expression. These expressions will be the components of our equation. Let's assemble our equation according to this diagram and the table above. That is, replace the words father, mother, son, daughter with their corresponding expression in the table:

age of father of mother of son and daughter equation 1

The expression corresponding to mom's age x - 3 was put in brackets for clarity.

Now let's solve the resulting equation. To begin with, you can open the brackets where you can:

age of father of mother of son and daughter equation 2

To free the equation from fractions, multiply both parts by 3

age of father of mother of son and daughter equation 3

Solve the resulting equation using the known identity transformations:

age of father of mother of son and daughter equation 4

We found the value of variable x. This variable was responsible for the age of the father. So the age of the father is 36 years old.

Knowing the age of the father, we can calculate the ages of the other family members. To do this, substitute the value of variable x in those expressions that correspond to the age of a particular family member.

The task said that my mother is 3 years younger than my father. We denoted her age by the expression x - 3. The value of the variable x is now known, and to calculate mom's age, substitute the found value 36 instead of x in the expression x - 3

x - 3 = 36 - 3 = 33 years old mom.

The age of the other family members is determined similarly:

age of the father of the mother of the son and daughter table 4

Task 4. A kilogram of caviar costs x dollars. Write an expression that calculates how many kilograms of caviar you can buy for $300.

300 * x

Task 5. x dollars was used to buy 5 kilograms of strawberries. Write an expression that calculates how many dollars one kilogram of strawberries costs.

x * 5

Task 6. Tom, John, and Leo went to the cafeteria during recess and bought a large pizza and a glass of fresh juice. The large pizza cost x dollars and the coffee cost $15. Determine the cost of the pizza if you know that $120 was paid for everything?

Of course, this task is simple and can be solved without resorting to an equation. To do this, subtract the cost of three glasses of juice from $120 (15 × 3), and divide the result by 3

the cost of sandwiches and coffee a simple solution

But our goal is to make an equation to the task and solve this equation. So, the cost of a large pizza is x dollars. Only three of them were bought. So if we increase the cost by three times, we get an expression describing how many dollars were paid for the three large pizzas.

3x — the cost of three large pizzas

And the cost of three glasses of juice can be written as 15 × 3. 15 is the cost of one glass of juice, and 3 is the multiplier (Tom, John, and Leo) that triples that cost.

Under the terms of the task, $120 is paid for everything. We already have an approximate scheme of what needs to be done:

Cost of three large pizzas + cost of three glasses of fresh juice = $120

The expressions describing the cost of three pizzas and three glasses of juice are ready. These are expressions 3x and 15 × 3. Use the scheme to make an equation and solve it:

cost of sandwiches and coffee solution to the problem

So, the cost of one large pizza is $25.

The task is solved correctly only if the equation to it is written correctly. Unlike ordinary equations by which we learn to find the roots, problem-solving equations have their own specific application. Each component of such an equation can be described in verbal form. When making an equation, it is imperative that we understand why we are including one component or another and why it is needed.

You must also remember that after solving the equation, the left-hand side will have to equal the right-hand side. The equation composed must not contradict this idea.

Let's imagine that the equation is a scale with two bowls and a screen showing the state of the scale.

scales

At this point, the screen shows an equal sign. It is clear why the left bowl is equal to the right bowl - there is nothing on the scales. The state of the scales and the absence of anything on the bowls will be recorded with the following equality:

Put a watermelon on the left side of the scale:

watermelon scales on the left bowl

The left bowl outweighed the right bowl and the screen sounded an alarm, showing a not equal sign ( ≠ ). This sign indicates that the left bowl is not equal to the right bowl.

Now let's try to solve the problem. Let's find out how much the watermelon on the left bowl weighs. But how can we find this out? Our scales are only for checking if the left bowl is equal to the right one.

Equations come to the rescue. Recall that an equation contains a variable whose value must be found. The scale in this case is the equation itself, and the mass of the watermelon is the variable, the value of which needs to be found. Our goal is to get this equation right. That is, to align the scales so that we can calculate the mass of the watermelon.

To level the scales, we can put some heavy object on the right-hand bowl. For example, let's put a weight of 7 kilograms there.

watermelon scales on the left side and a 7 kg weight on the right side

Now the right bowl outweighs the left bowl. The screen still shows that the bowls are not equal.

Let's try to put a weight of 4 kg on the left bowl

Watermelon on the left-hand scale and a 4 kg weight on the right-hand scale and a 7 kg weight on the right-hand scale

Now the scales are aligned. In the picture you can see that the left bowl is at the level of the right bowl. And the screen shows an equal sign. This sign says that the left bowl is equal to the right bowl.

So we got an equation - an equality that contains the unknown. The left bowl is the left side of the equation, consisting of the components 4 and the variable x (watermelon mass), and the right bowl is the right side of the equation, consisting of the component 7.

Watermelon on the left-hand scale and a 4 kg weight on the right-hand scale and a 7 kg weight on the right-hand scale

Well, it's not hard to guess that the root of the equation 4 + x = 7 is 3. So the mass of the watermelon is 3 kg.

The same is true for the other tasks. To find some unknown value, various elements are added to the left or right side of the equation: terms, multipliers, expressions. In school tasks, these elements may already be given. All that remains is to structure them correctly and construct the equation. In this example, we were trying weights of different weights to calculate the mass of the watermelon.

Naturally, the data given in the task must first be reduced to a form in which they can be included in the equation.

Consider the following task. The father's age is equal to the age of the son and daughter together. The son is twice as old as the daughter and twenty years younger than the father. How old are each?

The daughter's age can be denoted by x. If the son is twice as old as the daughter, his age will be denoted by 2x. The condition of the problem says that together the age of the daughter and the son are equal to the age of the father. So the father's age will be denoted by the sum x + 2x

father and son and daughter age scales table

In the expression x + 2x you can give like terms. Then the age of the father will be denoted as 3x

Now let's make an equation. We need to get an equation where we can find the unknown x. Let's use weights. On the left hand side put the age of the father (3x), and on the right hand side put the age of the son (2x)

father and son age scales

It is clear why the left bowl outweighs the right and why the screen shows the sign ( ≠ ). After all, it is logical that the age of the father is greater than the age of the son.

But we need to equalize the scales to be able to calculate the unknown x. To do this we need to add some number to the right-hand scale. What exact number is given in the task. The condition said that the son is 20 years younger than his father. So 20 years is the number to put on the scale.

The scales will even out if we add these 20 years to the right side of the scale. In other words, raise the son to the age of the father.

scales father's and son's age plus 20 years on the left hand side

Now the scales are aligned. We got the equation 3x = 2x + 20, which is easy to solve:

2x + 20 = 3x Solution

At the beginning of this task, we used the variable x to represent our daughter's age. Now we found the value of this variable. The daughter is 20 years old.

Next, it was said that the son is two years older than the daughter, so the son is (20 × 2), that is, 40 years old.

Finally, let's calculate the age of the father. It was said in the problem that he is equal to the sum of the ages of the son and daughter, that is, (20 + 40) years old.

2x + 20 = 3x Solution

Let's return to the middle of the task and note one point. When we put the age of the father and the age of the son on the scale, the left cup outweighed the right

But we solved this problem by adding another 20 years to the right-hand scale. As a result, the scales aligned and we got the equality 3x = 2x + 20

more problem solving using equations

But we could not add those 20 years to the right cup, but subtract them from the left. We would have obtained equality and in that case

scales father and son age minus 20 years on the other side

This time you get the equation 3x -20 = 2x. The root of the equation is still 20

3x - 20 = 2x Solution

That is, the equations 3x = 2x + 20 and 3x -20 = 2x are equipotent. And we remember that equal equations have the same roots. If you look closely at these two equations, you can see that the second equation is obtained by moving the number 20 from the right side to the left side with the opposite sign. And this action, as stated in the previous lesson, does not change the roots of the equation.

Also note that at the beginning of the task, the ages of each family member could be given by other expressions.

x second

Let's solve this equation

x plus 20 = x plus x * 2 Solution

In other words, the task can be solved by different methods. So you should not be discouraged if you cannot solve a particular problem. But you should keep in mind that there are the easiest ways to solve the problem. It is possible to get to the city center by different routes, but there is always the most convenient, fastest and safest route.

Examples of problem solving

Task 1. There are a total of 30 notebooks in two packs. If two notebooks were moved from the first stack to the second stack, the first stack would contain twice as many notebooks as the second stack. How many notebooks were in each pack?

Denote by x the number of notebooks that were in the first stack. If the total number of notebooks was 30, and the variable x is the number of notebooks in the first stack, then the number of notebooks in the second stack will be denoted by the expression 30 - x. That is, from the total number of notebooks we subtract the number of notebooks in the first stack, and thus we obtain the number of notebooks in the second stack.

a table of the number of notebooks in the first and second packs

Further it is said that if you move two notebooks from the first pack to the second pack, there will be twice as many notebooks in the first pack. So let's remove two notebooks from the first stack

the number of notebooks in the first and second packs we construct equation 2

and add these two notebooks to the second pack

the number of notebooks in the first and second packs we construct equation 3

The expressions from which we will compose the equation now take the following form:

table 2 number of notebooks in the first and second packs

Let's try to make an equation out of the available expressions. Put both stacks of notebooks on the scales

scales number of notebooks in the first and second pack

The left bowl is heavier than the right one. This is because the problem statement says that after taking two notebooks from the first stack and putting them into the second stack, the number of notebooks in the first stack became twice as many as in the second stack.

To equalize the scales and get the equation, let's double the right-hand side. To do this, multiply it by 2

scales number of notebooks in the first and second pack of Fig. 2

We obtain the equation x-2 = 2(30 - x +2) . Solve this equation:

x minus 2 = 2 * 0 minus x plus 2 Solution

We denoted the first packet by the variable x. Now we have found its value. The variable x is 22. So there were 22 notebooks in the first stack.

And we denoted the second tutu by the expression 30 - x, and since the value of the variable x is now known, we can calculate the number of notebooks in the second tutu. It is equal to 30 - 22, i.e. 8 pieces.

Task 2. Two people were peeling potatoes. One peeled two potatoes per minute and the other peeled three potatoes. Together they peeled 400 pieces. How long did each work, if the second worked 25 minutes longer than the first?

Denote by x the time the first person worked. Since the second person worked 25 minutes longer than the first person, his time will be denoted by x + 25

The first worker peeled 2 potatoes per minute, and since he worked x minutes, he peeled a total of 2x potatoes.

The second person peeled three potatoes per minute, and since he worked x + 25 minutes, he peeled a total of 3(x + 25) potatoes.

Together they peeled 400 potatoes

table two people peeled potatoes

From the available components let us make and solve the equation. The left side of the equation will be the potatoes peeled by each person, and the right side will be their sum:

2x plus 3x plus 75 = 400 Solution

At the beginning of this task, we used the variable x to denote the working time of the first person. Now we found the value of this variable. The first person worked for 65 minutes.

And the second person worked x + 25 minutes, and since the value of the variable x is now known, we can calculate the working time of the second person - it is 65 + 25, that is, 90 minutes.

The task from the Russian algebra textbook . (Their currency is rubles and kopecks (like dollars and cents))

Of the varieties of tea is a mixture of 32 kg. A kilogram of the first grade costs 8 rubles, and the second grade 6 rubles. 50 kop. How many kilos of both varieties, if a kilo of mixture costs (no profit and loss) 7 rubles. 10 kop.

Denote by x the mass of first grade tea. Then the mass of second grade tea will be denoted by the expression 32 - x

table 1 weight of first and second grade teas

A kilogram of first grade tea costs 8 rubles. If you multiply these eight rubles by the number of kilograms of first grade tea, you can find out how many rubles it cost x kilograms of first grade tea.

A kg of second class tea costs 6 roubles. 50 kopecks. If the price is 6 rubles. 50 kopecks times 32 - x, we will find out how much it cost 32 - x kg of tea of the second sort.

The condition says that a kilogram of mixture costs 7 rubles. 10 kopecks. All in all 32 kg of the mixture were produced. Multiply 7 rubles. 10 kopecks by 32 we can find out how much 32 kg of the mixture costs.

The expressions from which we will make the equation now take the following form:

Table 2 cost of first and second grade teas

Let's try to make an equation out of the available expressions. Let's put the cost of mixtures of first and second grade teas on the left scale, and on the right scale let's put the cost of 32 kg of mixture, i.e. the total cost of the mixture, which includes both varieties of tea:

scales the cost of first and second grade tea

We get the equation 8x + 6.50(32 - x) = 7.10 * 32 . Let's solve it:

8x plus 650 * 32 minus x = 710 * 32 Solution

At the beginning of this task, we used the variable x to denote the mass of the first grade tea. Now we have found the value of this variable. The variable x is 12.8. This means that 12.8 kg of first grade tea was used to make the mixture.

We have used the expression 32 - x to represent the mass of second-grade tea, and since the value of the variable x is now known, we can calculate the mass of second-grade tea. It is 32 - 12.8, i.e. 19.2. This means that 19.2 kg of second grade tea were used to prepare the mixture.

Seven point one two

Some tasks may involve topics that a person may not have studied. This task is one of those tasks. It touches on the concepts of distance, speed, and time. Accordingly, to solve such a problem, it is necessary to have an idea of the things mentioned in the problem. In our case, it is necessary to know what represents distance, speed and time.

In the task we need to find the distances of the two roads. We have to make an equation that will allow us to calculate these distances.

Let's remember how distance, speed, and time are interrelated. Each of these quantities can be described using a letter equation:

distance speed time in the picture

We will use the right side of one of these equations to make our own equation. To find out which one, go back to the text of the problem and pay attention to the following point:

t = s * v

Now let's make an equation out of the available expressions

scales two distances on the bowls

Through the variable S we denoted the distance of the first road. Now we have found the value of this variable. The variable S is 15. So the distance of the first road is 15 km.

And the distance of the second road is S + 3. Since the value of the variable S is now known, we can calculate the distance of the second road. This distance is equal to the sum of 15 + 3, that is 18 km.

Task 4. Two cars walk along the highway at the same speed. If the first increases the speed by 10 km/h, and the second decreases the speed by 10 km/h, then the first car will pass the same distance in 2 h as the second car did in 3 h. At what speed are the cars traveling?

Denote by v the speed of each car. Further in the problem there are hints: increase the speed of the first car by 10 km/h, and decrease the speed of the second car by 10 km/h. Let us use this hint

v plus 10 v minus 10

Further it is said that at such speeds (increased and decreased by 10 km/h) the first car will cover the same distance in 2 hours as the second car did in 3 hours. The phrase "as much" can be understood as "the distance traveled by the first car will be equal to the distance traveled by the second car" .

The distance, as we remember, is determined by the formula S = vt. We are interested in the right part of this letter equation - it will allow us to make an equation containing the variable v.

So, at speed v + 10 km/h, the first car will travel 2(v+10) km and the second car will travel 3(v - 10) km. Under this condition, the cars will travel the same distance, so it is sufficient to connect the two expressions with an equal sign to obtain the equation. Then we obtain the equation 2(v+10) = 3(v - 10). Solve it:

2v plus 20 = 3v minus 30 step 1

In the task condition it was said that the cars go at the same speed. We denoted this speed by the variable v. Now we have found the value of this variable. The variable v is 50. So the speed of both cars was 50 km/h.

Task 5. In 9 hours along the river the ship travels the same distance as in 11 hours against the stream. Find the boat's own speed if the speed of the river flow is 2 km/h.

Denote by v the boat's own speed. The speed of the river flow is 2 km/h. The speed of the ship along the river will be v + 2 km/h, and against the current - (v - 2) km/h.

The statement of the task says that the boat takes the same distance in 9 hours upstream as it took 11 hours upstream. The phrase "the same distance" can be understood as "the distance traveled by the ship upstream in 9 hours is equal to the distance traveled by the ship against the river in 11 hours". That is, the distances will be the same.

The distance is determined by the formula S = vt. Let's use the right part of this letter equation to make our own equation.

So, the boat will travel 9(v + 2) km in 9 hours upstream, and 11(v - 2) km in 11 hours upstream. Since both expressions describe the same distance, let's equate the first expression to the second one. The resulting equation is 9(v + 2) = 11(v - 2) 

9v plus 18 = 11v minus 22 Solution

So the proper speed of the motorboat is 20 km/h.

When solving tasks, it is a useful habit to determine in advance on which set the solution is sought.

Suppose that in the task we needed to find the time in which the pedestrian travels the specified path. We denoted the time by the variable t, then made an equation containing this variable and found its value.

We know from practice that the time of motion of an object can take both integer and fractional values, for example 2 h, 1.5 h, and 0.5 h. Then we can say that the solution of this problem is searched for on the set of rational numbers Q, because each value of 2 h, 1.5 h, 0.5 h can be represented as a fraction.

So after you denote an unknown quantity by a variable, it is useful to specify which set this quantity belongs to. In our example, time t belongs to the set of rational numbers Q

You can also introduce a restriction for the variable t, stating that it can only take positive values. Indeed, if an object has spent a certain amount of time on the path, then this time cannot be negative. Therefore next to the expression t ∈ Q we specify that its value must be greater than zero:

t  ∈  R , t  > 0

If by solving the equation we get a negative value for the variable t, then we can conclude that the problem is solved incorrectly, because this solution will not satisfy the condition t ∈ Q, t > 0.

Another example. If we were solving a task that required us to find the number of people to do a particular job, we would denote this number by the variable x. In such a task the solution would be searched for on the set of natural numbers

x  ∈ N

Indeed, the number of people is a whole number, e.g., 2 people, 3 people, 5 people. But not 1.5 (one whole person and half a person) or 2.3 (two whole persons and three tenths of a person).

Here we could specify that the number of people must be greater than zero, but the numbers included in the set of natural numbers N are themselves positive and greater than zero. In this set there are no negative numbers and no number 0. Therefore the expression x > 0 can be omitted. 

Task 6. A team of 2.5 times as many painters as carpenters arrived to repair a school. Soon the foreman added four more painters to the brigade, and transferred two carpenters to another site. As a result, there were four times as many painters as carpenters in the brigade. How many painters and how many carpenters were in the brigade originally

Denote by x the carpenters who arrived initially for repairs.

The number of carpenters is an integer greater than zero. Therefore let us specify that x belongs to the set of natural numbers

x  ∈ N

There were 2.5 times more painters than carpenters. Therefore, the number of painters will be denoted as 2.5x.

the number of carpenters x and the number of painters is two and a half times larger

It goes on to say that the foreman included four more painters in the crew, and transferred two carpenters to another site. Let's do the same for our expressions. Reduce the number of carpenters by 2

The number of carpenters is reduced by 2

And the number of painters will increase by 4

The number of carpenters is reduced by 4

Now the number of carpenters and painters will be denoted by the following expressions:

table new number of carpenters and painters

Let's try to make an equation out of the available expressions:

scales number of carpenters and painters

The right bowl is larger because after adding four more painters to the brigade and moving two carpenters to another site, the number of painters in the brigade is 4 times more than the number of carpenters. To equalize the scales, you need to quadruple the left bowl:

scales number of carpenters and painters equal bowls

The variable x was used to represent the original number of carpenters. Now we have found the value of this variable.  The variable x is equal to 8. So there were 8 carpenters in the team originally.

The number of painters was expressed as 2.5x, and since the value of the variable x is now known, we can calculate the number of painters, which is 2.5 × 8, that is, 20.

Go back to the beginning of the task and make sure that the condition x ∈ N is satisfied. The variable x is 8, and the elements of the set of natural numbers N are all numbers starting with 1, 2, 3 and so on to infinity. The same set includes the number 8, which we found.

8  ∈ N

The same can be said of the number of painters. The number 20 belongs to the set of natural numbers:

20  ∈ N

To understand the essence of the task and to correctly compose the equation, it is not necessary to use the model of scales with bowls. You can use other models: segments, tables, diagrams. You can come up with your own model, which would describe the essence of the task well.

Task 9. 30% of the milk was poured out of the can. This left 14 liters of milk in the can. How many liters of milk was in the can originally?

We need to find the original number of liters in the can. Let's represent the number of liters as a line and sign this line as X

x liters in the canister figure 1

It is said that 30% of the milk was poured out of the beaker. Let's draw an approximate figure of 30%

x liters in the canister figure 2

A percentage is, by definition, one hundredth of something. If 30% of the milk is poured, the other 70% is left in the can. This 70% is the 14 liters in the problem. Let's draw a picture of the remaining 70%

x liters in the canister figure 3

Now you can make an equation. Recall how to find the percentage of a number. To do this, the total amount of something is divided by 100 and the result is multiplied by the number of percentages you are looking for. Note that 14 liters, which is 70 percent, can be obtained in the same way: the original number of liters X divided by 100 and the result multiplied by 70. Equate all this to the number 14

x * 100 * 70 = 14

Or get a simpler equation: write 70% as 0.70, then multiply by X and equate that expression to 14

x * 100 * 70 = 14 otherwise Solution

So originally there were 20 liters of milk in the can.

Task 9. Two alloys of gold and silver are taken. One has the quantity of these metals in the ratio of 1 : 9, and the other 2 : 3. How much of each alloy do you need to take to get 15 kg of a new alloy in which the gold and silver would be 1 : 4?

Let's first try to find out how much gold and silver will be contained in 15 kg of the new alloy. The task says that the content of these metals should be 1 : 4, that is, one part of the alloy should contain gold, and four parts should contain silver. Then the total number of parts in the alloy will be 1 + 4 = 5, and the mass of one part will be 15 : 5 = 3 kg.

Let's determine how much gold is contained in 15 kg of the alloy. To do this, multiply 3 kg by the number of parts of gold:

3 kg × 1 = 3 kg

Let's determine how much silver will be contained in 15 kg of alloy:

3 kg × 4 = 12 kg

So an alloy with a mass of 15 kg will contain 3 kg of gold and 12 kg of silver. Now let's go back to the original alloys. We have to use each of them. Denote by x the mass of the first alloy, and the mass of the second alloy can be denoted by 15 - x

plate x and 15 minus x

Express in percentages all the ratios given in the problem and fill in the following table with them:

table three alloys figure 1

Let's transfer these data to the table. Let's enter 10% in the first line in the column "percentage of gold in the alloy" , 90% in the first line in the column "percentage of silver in the alloy" , and in the last column "mass of the alloy" we will enter the variable x, because that is how we designated the mass of the first alloy:

table three alloys figure 2

Let's transfer these data to the table. Let's enter 40% in the second line in the column "percentage of gold in the alloy", 60% in the second line in the column "percentage of silver in the alloy", and in the last column "mass of the alloy" let's enter the expression 15 - x, because that is how we denoted the mass of the second alloy:

table three alloys figure 3

Now you can use this table to make equations. Recall the tasks of concentration, alloys, and mixtures . If we separately add up the gold of both alloys and equate this sum to the mass of gold of the resulting alloy, we can find out what the value of x equals.

Further, for convenience we will express percentages in decimals.

The first alloy had 0.10x gold and the second alloy had 0.40(15 - x) gold. Then the mass of gold in the resulting alloy will be the sum of the masses of gold of the first and second alloys and this mass is 20% of the new alloy. And 20% of the new alloy is 3 kg of gold, calculated earlier. The resulting equation is 0.10x + 0.40(15 - x) = 3 . Let us solve this equation:

010x plus 040 * 15 - x = 3 Solution

Initially we denoted by x the mass of the first alloy. Now we have found the value of this variable. The variable x is 10. And the mass of the second alloy we denoted by 15 - x, and since the value of the variable x is now known, we can calculate the mass of the second alloy, it is equal to 15 - 10 = 5 kg.

So in order to obtain a new alloy with a mass of 15 kg in which gold and silver would be 1 : 4, we would have to take 10 kg of the first alloy and 5 kg of the second alloy.

The equation could also be made using the second column of the resulting table. Then we would get the equation 0.90x + 0.60(15 - x) = 12 . The root of this equation is also 10.

090x plus 060 * 15 minus x = 12 Solution

Task 10. There is an ore from two layers with copper grades of 6% and 11%. How much poor ore must be taken to get 20 tons with a copper content of 8% when mixed with the rich ore?

Denote by x the mass of poor ore. Since 20 tons of ore are to be obtained, 20 - x will be taken from the rich ore. Since the copper content in the poor ore is 6%, x tons of ore will contain 0.06x tons of copper. The rich ore has a copper content of 11%, and 20 - x tons of rich ore will contain 0.11(20 - x) tons of copper.

The resulting 20 tons of ore should have a copper content of 8%. This means that 20 tons of ore will contain 20 × 0.08 = 1.6 tons of copper.

Add 0.06x and 0.11(20 - x) and equate the sum to 1.6. We obtain the equation 0.06x + 0.11(20 - x) = 1.6

006x plus 011 * 20 minus x = 16

Let's solve this equation:

006x plus 011 * 20 minus x = 16 Solution

So, to get 20 tons of ore with 8% copper content, you need to take 12 tons of poor ore. The rich ore will be taken 20 - 12 = 8 tons.

Task 11. By increasing the average speed from 250 to 300 meters/minute, the athlete began to run the distance 1 minute faster. What is the length of the distance?

The length of the course (or distance of the course) can be described by the following letter equation:

distance formula figure for the problem

Let's use the right side of this equation to make our own equation. Initially, the athlete ran the distance at a speed of 250 meters per minute. At this speed, the length of the course would be described by the expression 250t

Then, the athlete increased her speed to 300 metres per minute. At this speed, the length of the course would be described by the expression 300t

Note that the length of the course is a constant. Whether the athlete increases her speed or decreases it, the length of the distance will remain the same.

This allows us to equate 250t with 300t, since both describe the same length.

250 t = 300 t

But the problem says that at a speed of 300 meters per minute, the athlete began to run the distance 1 minute faster. In other words, at 300 meters per minute, the running time will decrease by one. Therefore, in the equation 250t = 300t on the right hand side, the time should be reduced by one:

250t = 300 t - 1 Solution

We get the simplest equation. Let's solve it:

250t = 300 t - 1 Solution

At a speed of 250 meters per minute, the athlete runs the distance in 6 minutes. Knowing the speed and time, you can determine the length of the course:

S = 250 × 6 = 1500 m

And at 300 meters per minute, the athlete runs the distance in t - 1, that is, in 5 minutes. As stated earlier, the length of the distance does not change:

S  = 300 × 5 = 1500 m

Task 12. A rider catches up with a pedestrian 15 km ahead of him. In how many hours will the rider catch up with the pedestrian, if the first rides 10 km every hour and the second rides only 4 km?

This task is a motion task . It can be solved by determining the speed of approach and dividing the initial distance between the rider and the pedestrian by that speed.

The speed of convergence is determined by subtracting the slower speed from the faster speed:

10 km/h - 4 km/h = 6 km/h (approach speed)

With each hour, the distance of 15 kilometers will decrease by 6 kilometers. To find out when it is completely shortened (when the rider catches up with the pedestrian), divide 15 by 6

15 : 6 = 2,5 h

2.5 hours is two whole hours and a half hour. And half an hour is 30 minutes. So the rider will catch up with the pedestrian in two hours and 30 minutes.

rider catches up with pedestrian figure 1

Let's solve this problem using an equation.

Assume that the pedestrian and the rider set out from the same place. The pedestrian left before the rider and managed to cover 15 km

rider catches up with pedestrian figure 2

After that, the rider followed him on the road at a speed of 10 km/h. The speed of the pedestrian is only 4 km/h. This means that in some time the rider will catch up with the pedestrian. We need to find this time.

When the rider catches up with the pedestrian it will mean that they have traveled the same distance together. The distance traveled by the rider and the pedestrian is described by the following equation:

Let us use the right side of this equation to make our own equation.

The distance traveled by the rider will be described by the expression 10t. Since the pedestrian went ahead of the rider and managed to cover 15 km, the distance traveled by him will be described by the expression 4t + 15.

By the time the rider catches up with the pedestrian, they have both traveled the same distance. This allows us to equate the distances traveled by the rider and the pedestrian:

10t = 4t plus 15

Tasks for independent decision

Train speeds in this problem are measured in kilometers per hour. Therefore, convert the 45 min specified in the problem to hours. 45 min is 0.75 h

more problem solving using equations

Denote the time it takes for the freight train to arrive in the city by the variable t . Since the passenger train arrives in this city 0.75 h faster, its travel time will be denoted by the expression t −  0,75

The passenger train covered 48(t-0.75) km, and the freight train 36t km. Since we are talking about the same distance, let us equate the first expression with the second one. The resulting equation is 48(t - 0.75) = 36t. Solve it:

more problem solving using equations

Now calculate the distance between the cities. To do this, multiply the speed of the freight train (36 km/h) by its travel time t. The value of the variable t is now known - it is equal to three hours

36 × 3 = 108 km

You can also use the speed of the passenger train to calculate the distance. But in this case the value of the variable t should be reduced by 0.75 because the passenger train spent 0.75 h less time

48 × (3 − 0,75) = 144 − 36 = 108 km

Answer: The distance between the cities is 108 km.

Let t be the time after which the cars meet. Then the first car at the time of the meeting will have traveled 65t km, and the second 60t km. Add these distances and equate them to 150. We get the equation 65t + 60t = 150

more problem solving using equations

The value of the variable t is 1.2. So the cars met after 1.2 hours.

Answer: the cars met after 1.2 hours.

Let x be the number of workers in the first workshop. The second workshop had three times as many workers as the first workshop, so the number of workers in the second workshop can be denoted by the expression 3x. The third workshop had 15 fewer workers than the second workshop. Therefore, the number of workers in the third workshop can be denoted by the expression 3x - 15.

The task says that the total number of workers was 685. Therefore we can add the expressions x, 3x, 3x - 15 and equate that sum to the number 685. The result is x + 3x + (3x - 15) = 685

more problem solving using equations

The variable x was used to denote the number of workers in the first workshop. Now we have found the value of this variable, it is 100. So there were 100 workers in the first workshop.

The second workshop had 3x workers, so 3 × 100 = 300. And the third workshop had 3x - 15, so 3 × 100 - 15 = 285

Answer: In the first shop there were 100 workers, in the second - 300, in the third - 285.

Let x motors be repaired by the first workshop. Then the second workshop had to repair 18 - x motors.

Since the first workshop fulfilled its plan by 120%, that means it repaired 1.2x motors . And the second repair shop fulfilled its plan by 125%, so it repaired 1.25(18 - x) motors.

The problem says that 22 motors were repaired. Therefore we can add the expressions 1.2x and 1.25(18 - x) , then equate that sum to the number 22. The resulting equation is 1.2x + 1.25(18 - x) = 22

more problem solving using equations

The variable x was used to denote the number of motors to be repaired by the first workshop. Now we have found the value of this variable, it is 10. So the first workshop had to repair 10 motors.

The expression 18 - x represents the number of motors to be repaired by the second workshop. So the second workshop had to repair 18 - 10 = 8 motors.

Answer: The first workshop was to repair 10 engines and the second workshop was to repair 8 engines..

Let x dollars was the price of the product before the price increase. If the price increased by 30%, it means that it increased by $0.30x dollars. After the price increase, the product began to cost $91. Add x to 0.30x and equate that sum to 91. As a result we get the equation x + 0.30x = 91

more problem solving using equations

So before the price increase, the product cost $70.

Answer: Before the price increase, the product cost $70.

Let x be the initial number. Increase it by 25%. We obtain the expression x + 0.25x. Given the like terms, we obtain x + 0.25x = 1.25x.

Find what part of the original number x is from the new number 1.25x

more problem solving using equations

If the new number 1.25x is 100%, and the original number x is 80% of it, then reducing the new number by 20% you can get the original number x>

more problem solving using equations

Answer: to get the original number, the new number must be reduced by 20%.

Let x be the original number. Increase it by 20%. We obtain the expression x + 0.20x. Equate this sum to the number 144, so we obtain the equation x + 0.20x = 144

more problem solving using equations

Answer: the original value of the number is 120.

Let x be the original number. Decrease it by 10%. We obtain the expression x - 0.10x. Equate this difference to the number 45, so we get the equation x - 0.10x = 45

more problem solving using equations

Answer: the original value of the number is 50.

Let x be the original price of the album. Decrease this price by 15%, we get x - 0.15x. Reduce the price by another $15, so we get x - 0.15x - 15. After these reductions, the album now costs $19. Equate the expression x - 0.15x - 15 to the number 19, we get the equation x - 0.15x - 15 = 19

more problem solving using equations

Answer: The original album price is $40.

If 80% of the mass is lost, the remaining 20% will account for 4 tons of hay. Let x tons of grass be required to produce 4 tons of hay. If 4 tons will account for 20% of the grass, then we can make the equation:

more problem solving using equations

Answer: to get 4 tons of hay, you need to cut 20 tons of grass.

Let x kg of 20% salt solution be added to 1 kg of 10% salt solution.

1 kg of 10% salt solution contains 0.1 kg of salt. And x kg of 20% salt solution contains 0.20x kg of salt.

After adding x kg of the 20% solution, the new solution will contain 0.12(1 + x) kg of salt. We add 0.1 and 0.20x, then equate that sum to 0.12(1 + x). The resulting equation is 0.1 + 0.20x = 0.12(1 + x)

more problem solving using equations

Answer: to get a 12% salt solution, you need to add 0.25 kg of 20% solution to 1 kg of 10% solution.

Let x kg of the first solution be taken. Since 25 kg of solution must be prepared, the mass of the second solution can be denoted by the expression 25 - x.

The first solution will contain 0.20x kg of salt, and the second solution will contain 0.30(25 - x) kg of salt. The resulting solution contains 25 × 0.252 = 6.3 kg of salt. Add the expressions 0.20x and 0.30(25 - x), then equate that sum to 6.3. The resulting equation is

more problem solving using equations

So the first solution should take 12 kg, and the second 25 - 12 = 13 kg.

Answer: the first solution should take 12 kg, and the second 13 kg.

  • Basic theory of equation
  • Proportions: exercises

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CHAPTER 5 Solving First Degree Equations in One Variable

5.7 Use a Problem-Solving Strategy

Learning Objectives

By the end of this section, you will be able to:

  • Approach word problems with a positive attitude
  • Use a problem-solving strategy for word problems
  • Solve number problems

Approach Word Problems with a Positive Attitude

“If you think you can… or think you can’t… you’re right.”—Henry Ford

The world is full of word problems! Will my income qualify me to rent that apartment? How much punch do I need to make for the party? What size diamond can I afford to buy my girlfriend? Should I fly or drive to my family reunion?

How much money do I need to fill the car with gas? How much tip should I leave at a restaurant? How many socks should I pack for vacation? What size turkey do I need to buy for Thanksgiving dinner, and then what time do I need to put it in the oven? If my sister and I buy our mother a present, how much does each of us pay?

Now that we can solve equations, we are ready to apply our new skills to word problems. Do you know anyone who has had negative experiences in the past with word problems? Have you ever had thoughts like the student below?

A student is shown with thought bubbles saying “I don’t know whether to add, subtract, multiply, or divide!,” “I don’t understand word problems!,” “My teachers never explained this!,” “If I just skip all the word problems, I can probably still pass the class,” and “I just can’t do this!”

When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. We need to calm our fears and change our negative feelings.

Start with a fresh slate and begin to think positive thoughts. If we take control and believe we can be successful, we will be able to master word problems! Read the positive thoughts in (Figure 2) and say them out loud.

A student is shown with thought bubbles saying “While word problems were hard in the past, I think I can try them now,” “I am better prepared now. I think I will begin to understand word problems,” “I think I can! I think I can!,” and “It may take time, but I can begin to solve word problems.”

Think of something, outside of school, that you can do now but couldn’t do 3 years ago. Is it driving a car? Snowboarding? Cooking a gourmet meal? Speaking a new language? Your past experiences with word problems happened when you were younger—now you’re older and ready to succeed!

Use a Problem-Solving Strategy for Word Problems

We have reviewed translating English phrases into algebraic expressions, using some basic mathematical vocabulary and symbols. We have also translated English sentences into algebraic equations and solved some word problems. The word problems applied math to everyday situations. We restated the situation in one sentence, assigned a variable, and then wrote an equation to solve the problem. This method works as long as the situation is familiar and the math is not too complicated.

Now, we’ll expand our strategy so we can use it to successfully solve any word problem. We’ll list the strategy here, and then we’ll use it to solve some problems. We summarize below an effective strategy for problem solving.

Use a Problem-Solving Strategy to Solve Word Problems.

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose a variable to represent that quantity.
  • Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
  • Solve the equation using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

Pilar bought a purse on sale for $18, which is one-half of the original price. What was the original price of the purse?

Step 1. Read the problem. Read the problem two or more times if necessary. Look up any unfamiliar words in a dictionary or on the internet.

  • In this problem, is it clear what is being discussed? Is every word familiar?

Step 2. Identify what you are looking for. Did you ever go into your bedroom to get something and then forget what you were looking for? It’s hard to find something if you are not sure what it is! Read the problem again and look for words that tell you what you are looking for!

  • In this problem, the words “what was the original price of the purse” tell us what we need to find.

Step 3. Name what we are looking for. Choose a variable to represent that quantity. We can use any letter for the variable, but choose one that makes it easy to remember what it represents.

p=

Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Translate the English sentence into an algebraic equation.

Reread the problem carefully to see how the given information is related. Often, there is one sentence that gives this information, or it may help to write one sentence with all the important information. Look for clue words to help translate the sentence into algebra. Translate the sentence into an equation.

Step 5. Solve the equation using good algebraic techniques. Even if you know the solution right away, using good algebraic techniques here will better prepare you to solve problems that do not have obvious answers.

p=36

  • Does $36 make sense in the problem? Yes, because 18 is one-half of 36, and the purse was on sale at half the original price.

Step 7. Answer the question with a complete sentence. The problem asked “What was the original price of the purse?”

  • The answer to the question is: “The original price of the purse was $36.”

If this were a homework exercise, our work might look like this:

Pilar bought a purse on sale for $18, which is one-half the original price. What was the original price of the purse?

Joaquin bought a bookcase on sale for $120, which was two-thirds of the original price. What was the original price of the bookcase?

Two-fifths of the songs in Mariel’s playlist are country. If there are 16 country songs, what is the total number of songs in the playlist?

Let’s try this approach with another example.

Ginny and her classmates formed a study group. The number of girls in the study group was three more than twice the number of boys. There were 11 girls in the study group. How many boys were in the study group?

Guillermo bought textbooks and notebooks at the bookstore. The number of textbooks was 3 more than twice the number of notebooks. He bought 7 textbooks. How many notebooks did he buy?

Gerry worked Sudoku puzzles and crossword puzzles this week. The number of Sudoku puzzles he completed is eight more than twice the number of crossword puzzles. He completed 22 Sudoku puzzles. How many crossword puzzles did he do?

Solve Number Problems

Now that we have a problem solving strategy, we will use it on several different types of word problems. The first type we will work on is “number problems.” Number problems give some clues about one or more numbers. We use these clues to write an equation. Number problems don’t usually arise on an everyday basis, but they provide a good introduction to practicing the problem solving strategy outlined above.

The difference of a number and six is 13. Find the number.

The difference of a number and eight is 17. Find the number.

-7

The sum of twice a number and seven is 15. Find the number.

Did you notice that we left out some of the steps as we solved this equation? If you’re not yet ready to leave out these steps, write down as many as you need.

The sum of four times a number and two is 14. Find the number.

The sum of three times a number and seven is 25. Find the number.

Some number word problems ask us to find two or more numbers. It may be tempting to name them all with different variables, but so far we have only solved equations with one variable. In order to avoid using more than one variable, we will define the numbers in terms of the same variable. Be sure to read the problem carefully to discover how all the numbers relate to each other.

One number is five more than another. The sum of the numbers is 21. Find the numbers.

One number is six more than another. The sum of the numbers is twenty-four. Find the numbers.

The sum of two numbers is fifty-eight. One number is four more than the other. Find the numbers.

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

The sum of two numbers is negative twenty-three. One number is seven less than the other. Find the numbers.

-15,-8

One number is ten more than twice another. Their sum is one. Find the numbers.

One number is eight more than twice another. Their sum is negative four. Find the numbers.

-4,0

Some number problems involve consecutive integers. Consecutive integers are integers that immediately follow each other.

 Examples of consecutive integers are:

\begin{array}{c}1,2,3,4\hfill \\ \\ -10,-9,-8,-7\hfill \\ 150,151,152,153\hfill \end{array}

The sum of two consecutive integers is 47. Find the numbers.

95

Now that we have worked with consecutive integers, we will expand our work to include consecutive even integers and consecutive odd integers. Consecutive even integers are even integers that immediately follow one another. Examples of consecutive even integers are:

\begin{array}{c}18,20,22\hfill \\ \\ 64,66,68\hfill \\ -12,-10,-8\hfill \end{array}

Consecutive odd integers are odd integers that immediately follow one another. Consider the consecutive odd integers 77, 79, and 81

\begin{array}{c}77,79,81\hfill \\ \\ n,n+2,n+4\hfill \end{array}

Does it seem strange to add 2 (an even number) to get from one odd integer to the next? Do you get an odd number or an even number when we add 2 to 3? to 11? to 47?

Whether the problem asks for consecutive even numbers or odd numbers, you don’t have to do anything different. The pattern is still the same—to get from one odd or one even integer to the next, add 2

Find three consecutive even integers whose sum is 84

TRY IT 10.1

Find three consecutive even integers whose sum is 102

TRY IT 10.2

-24

A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

TRY IT 11.1

According to the National Automobile Dealers Association, the average cost of a car in 2014 was 28,500. This was 1,500 less than 6 times the cost in 1975. What was the average cost of a car in 1975?

TRY IT 11.2

The Canadian Real Estate Association (CREA) data shows that the median price of new home in the Canada in December 2018 was $470,000. This was $14,000 more than 19 times the price in December 1967. What was the median price of a new home in December 1967?

$24,000

Key Concepts

  • Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.

\begin{array}{cccc}\hfill n\hfill & & & {1}^{\text{st}}\phantom{\rule{0.2em}{0ex}}\text{integer}\hfill \\ \hfill n+1\hfill & & & {2}^{\text{nd}}\phantom{\rule{0.2em}{0ex}}\text{integer consecutive integer}\hfill \\ \hfill n+2\hfill & & & {3}^{\text{rd}}\phantom{\rule{0.2em}{0ex}}\text{consecutive integer . . . etc.}\hfill \end{array}

Consecutive even integers are even integers that immediately follow one another.

\begin{array}{cccc}\hfill n\hfill & & & {1}^{\text{st}}\phantom{\rule{0.2em}{0ex}}\text{integer}\hfill \\ \hfill n+2\hfill & & & {2}^{\text{nd}}\phantom{\rule{0.2em}{0ex}}\text{integer consecutive integer}\hfill \\ \hfill n+4\hfill & & & {3}^{\text{rd}}\phantom{\rule{0.2em}{0ex}}\text{consecutive integer . . . etc.}\hfill \end{array}

Consecutive odd integers are odd integers that immediately follow one another.

Practice Makes Perfect

Use the approach word problems with a positive attitude.

In the following exercises, prepare the lists described.

In the following exercises, solve using the problem solving strategy for word problems. Remember to write a complete sentence to answer each question.

In the following exercises, solve each number word problem.

Everyday Math

Writing exercises, attributions.

This chapter has been adapted from “Use a Problem-Solving Strategy” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence . Adapted by Izabela Mazur. See the Copyright page for more information.

Introductory Algebra by Izabela Mazur is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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  • 2.3 Models and Applications
  • Introduction to Prerequisites
  • 1.1 Real Numbers: Algebra Essentials
  • 1.2 Exponents and Scientific Notation
  • 1.3 Radicals and Rational Exponents
  • 1.4 Polynomials
  • 1.5 Factoring Polynomials
  • 1.6 Rational Expressions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Equations and Inequalities
  • 2.1 The Rectangular Coordinate Systems and Graphs
  • 2.2 Linear Equations in One Variable
  • 2.4 Complex Numbers
  • 2.5 Quadratic Equations
  • 2.6 Other Types of Equations
  • 2.7 Linear Inequalities and Absolute Value Inequalities
  • Introduction to Functions
  • 3.1 Functions and Function Notation
  • 3.2 Domain and Range
  • 3.3 Rates of Change and Behavior of Graphs
  • 3.4 Composition of Functions
  • 3.5 Transformation of Functions
  • 3.6 Absolute Value Functions
  • 3.7 Inverse Functions
  • Introduction to Linear Functions
  • 4.1 Linear Functions
  • 4.2 Modeling with Linear Functions
  • 4.3 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 5.1 Quadratic Functions
  • 5.2 Power Functions and Polynomial Functions
  • 5.3 Graphs of Polynomial Functions
  • 5.4 Dividing Polynomials
  • 5.5 Zeros of Polynomial Functions
  • 5.6 Rational Functions
  • 5.7 Inverses and Radical Functions
  • 5.8 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 6.1 Exponential Functions
  • 6.2 Graphs of Exponential Functions
  • 6.3 Logarithmic Functions
  • 6.4 Graphs of Logarithmic Functions
  • 6.5 Logarithmic Properties
  • 6.6 Exponential and Logarithmic Equations
  • 6.7 Exponential and Logarithmic Models
  • 6.8 Fitting Exponential Models to Data
  • Introduction to Systems of Equations and Inequalities
  • 7.1 Systems of Linear Equations: Two Variables
  • 7.2 Systems of Linear Equations: Three Variables
  • 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 7.4 Partial Fractions
  • 7.5 Matrices and Matrix Operations
  • 7.6 Solving Systems with Gaussian Elimination
  • 7.7 Solving Systems with Inverses
  • 7.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 8.1 The Ellipse
  • 8.2 The Hyperbola
  • 8.3 The Parabola
  • 8.4 Rotation of Axes
  • 8.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 9.1 Sequences and Their Notations
  • 9.2 Arithmetic Sequences
  • 9.3 Geometric Sequences
  • 9.4 Series and Their Notations
  • 9.5 Counting Principles
  • 9.6 Binomial Theorem
  • 9.7 Probability

Learning Objectives

In this section, you will:

  • Set up a linear equation to solve a real-world application.
  • Use a formula to solve a real-world application.

Neka is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum number of points that can be earned is 100. Is it possible for Neka to end the course with an A? A simple linear equation will give Neka his answer.

Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.

Setting up a Linear Equation to Solve a Real-World Application

To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write 0.10 x . 0.10 x . This expression represents a variable cost because it changes according to the number of miles driven.

If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost C . C .

When dealing with real-world applications, there are certain expressions that we can translate directly into math. Table 1 lists some common verbal expressions and their equivalent mathematical expressions.

Given a real-world problem, model a linear equation to fit it.

  • Identify known quantities.
  • Assign a variable to represent the unknown quantity.
  • If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
  • Write an equation interpreting the words as mathematical operations.
  • Solve the equation. Be sure the solution can be explained in words, including the units of measure.

Modeling a Linear Equation to Solve an Unknown Number Problem

Find a linear equation to solve for the following unknown quantities: One number exceeds another number by 17 17 and their sum is 31. 31. Find the two numbers.

Let x x equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as x + 17. x + 17. The sum of the two numbers is 31. We usually interpret the word is as an equal sign.

The two numbers are 7 7 and 24. 24.

Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is 36 , 36 , find the numbers.

Setting Up a Linear Equation to Solve a Real-World Application

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time.

  • ⓐ Write a linear equation that models the packages offered by both companies.
  • ⓑ If the average number of minutes used each month is 1,160, which company offers the better plan?
  • ⓒ If the average number of minutes used each month is 420, which company offers the better plan?
  • ⓓ How many minutes of talk-time would yield equal monthly statements from both companies?
  • ⓐ The model for Company A can be written as A = 0.05 x + 34. A = 0.05 x + 34. This includes the variable cost of 0.05 x 0.05 x plus the monthly service charge of $34. Company B ’s package charges a higher monthly fee of $40, but a lower variable cost of 0.04 x . 0.04 x . Company B ’s model can be written as B = 0.04 x + $ 40. B = 0.04 x + $ 40.

If the average number of minutes used each month is 1,160, we have the following:

So, Company B offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A when the average number of minutes used each month is 1,160.

If the average number of minutes used each month is 420, we have the following:

If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B ’s monthly cost of $56.80.

To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of ( x , y ) ( x , y ) coordinates: At what point are both the x- value and the y- value equal? We can find this point by setting the equations equal to each other and solving for x.

Check the x- value in each equation.

Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See Figure 2

Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses?

Using a Formula to Solve a Real-World Application

Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, A = L W ; A = L W ; the perimeter of a rectangle, P = 2 L + 2 W ; P = 2 L + 2 W ; and the volume of a rectangular solid, V = L W H . V = L W H . When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.

Solving an Application Using a Formula

It takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work?

This is a distance problem, so we can use the formula d = r t , d = r t , where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.

First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or 1 2 1 2 h at rate r . r . His drive home takes 40 min, or 2 3 2 3 h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance d . d . A table, such as Table 2 , is often helpful for keeping track of information in these types of problems.

Write two equations, one for each trip.

As both equations equal the same distance, we set them equal to each other and solve for r .

We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d.

The distance between home and work is 20 mi.

Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for r . r .

On Saturday morning, it took Jennifer 3.6 h to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 h to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday?

Solving a Perimeter Problem

The perimeter of a rectangular outdoor patio is 54 54 ft. The length is 3 3 ft greater than the width. What are the dimensions of the patio?

The perimeter formula is standard: P = 2 L + 2 W . P = 2 L + 2 W . We have two unknown quantities, length and width. However, we can write the length in terms of the width as L = W + 3. L = W + 3. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in Figure 3 .

Now we can solve for the width and then calculate the length.

The dimensions are L = 15 L = 15 ft and W = 12 W = 12 ft.

Find the dimensions of a rectangle given that the perimeter is 110 110 cm and the length is 1 cm more than twice the width.

Solving an Area Problem

The perimeter of a tablet of graph paper is 48 in. The length is 6 6 in. more than the width. Find the area of the graph paper.

The standard formula for area is A = L W ; A = L W ; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.

We know that the length is 6 in. more than the width, so we can write length as L = W + 6. L = W + 6. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.

Now, we find the area given the dimensions of L = 15 L = 15 in. and W = 9 W = 9 in.

The area is 135 135 in. 2 .

A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft 2 of new carpeting should be ordered?

Solving a Volume Problem

Find the dimensions of a shipping box given that the length is twice the width, the height is 8 8 inches, and the volume is 1,600 in. 3 .

The formula for the volume of a box is given as V = L W H , V = L W H , the product of length, width, and height. We are given that L = 2 W , L = 2 W , and H = 8. H = 8. The volume is 1,600 1,600 cubic inches.

The dimensions are L = 20 L = 20 in., W = 10 W = 10 in., and H = 8 H = 8 in.

Note that the square root of W 2 W 2 would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.

Access these online resources for additional instruction and practice with models and applications of linear equations.

  • Problem solving using linear equations
  • Problem solving using equations
  • Finding the dimensions of area given the perimeter
  • Find the distance between the cities using the distance = rate * time formula
  • Linear equation application (Write a cost equation)

2.3 Section Exercises

To set up a model linear equation to fit real-world applications, what should always be the first step?

Use your own words to describe this equation where n is a number: 5 ( n + 3 ) = 2 n 5 ( n + 3 ) = 2 n

If the total amount of money you had to invest was $2,000 and you deposit x x amount in one investment, how can you represent the remaining amount?

If a carpenter sawed a 10-ft board into two sections and one section was n n ft long, how long would the other section be in terms of n n ?

If Bill was traveling v v mi/h, how would you represent Daemon’s speed if he was traveling 10 mi/h faster?

Real-World Applications

For the following exercises, use the information to find a linear algebraic equation model to use to answer the question being asked.

Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?

Beth and Ann are joking that their combined ages equal Sam’s age. If Beth is twice Ann’s age and Sam is 69 yr old, what are Beth and Ann’s ages?

Ruden originally filled out 8 more applications than Hanh. Then each boy filled out 3 additional applications, bringing the total to 28. How many applications did each boy originally fill out?

For the following exercises, use this scenario: Two different telephone carriers offer the following plans that a person is considering. Company A has a monthly fee of $20 and charges of $.05/min for calls. Company B has a monthly fee of $5 and charges $.10/min for calls.

Find the model of the total cost of Company A’s plan, using m m for the minutes.

Find the model of the total cost of Company B’s plan, using m m for the minutes.

Find out how many minutes of calling would make the two plans equal.

If the person makes a monthly average of 200 min of calls, which plan should for the person choose?

For the following exercises, use this scenario: A wireless carrier offers the following plans that a person is considering. The Family Plan: $90 monthly fee, unlimited talk and text on up to 8 lines, and data charges of $40 for each device for up to 2 GB of data per device. The Mobile Share Plan: $120 monthly fee for up to 10 devices, unlimited talk and text for all the lines, and data charges of $35 for each device up to a shared total of 10 GB of data. Use P P for the number of devices that need data plans as part of their cost.

Find the model of the total cost of the Family Plan.

Find the model of the total cost of the Mobile Share Plan.

Assuming they stay under their data limit, find the number of devices that would make the two plans equal in cost.

If a family has 3 smart phones, which plan should they choose?

For exercises 17 and 18, use this scenario: A retired woman has $50,000 to invest but needs to make $6,000 a year from the interest to meet certain living expenses. One bond investment pays 15% annual interest. The rest of it she wants to put in a CD that pays 7%.

If we let x x be the amount the woman invests in the 15% bond, how much will she be able to invest in the CD?

Set up and solve the equation for how much the woman should invest in each option to sustain a $6,000 annual return.

Two planes fly in opposite directions. One travels 450 mi/h and the other 550 mi/h. How long will it take before they are 4,000 mi apart?

Ben starts walking along a path at 4 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the same path at 6 mi/h. How long will it be before Amanda catches up to Ben?

Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 70 mi takes her 4.5 h. For what distance did she travel at 20 mi/h?

A chemistry teacher needs to mix a 30% salt solution with a 70% salt solution to make 20 qt of a 40% salt solution. How many quarts of each solution should the teacher mix to get the desired result?

Raúl has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest and part at 12% interest. How much does Raúl need to invest in each option to make get a total 11% return on his $20,000?

For the following exercises, use this scenario: A truck rental agency offers two kinds of plans. Plan A charges $75/wk plus $.10/mi driven. Plan B charges $100/wk plus $.05/mi driven.

Write the model equation for the cost of renting a truck with plan A.

Write the model equation for the cost of renting a truck with plan B.

Find the number of miles that would generate the same cost for both plans.

If Tim knows he has to travel 300 mi, which plan should he choose?

For the following exercises, use the formula given to solve for the required value.

A = P ( 1 + r t ) A = P ( 1 + r t ) is used to find the principal amount P deposited, earning r % interest, for t years. Use this to find what principal amount P David invested at a 3% rate for 20 yr if A = $ 8,000. A = $ 8,000.

The formula F = m v 2 R F = m v 2 R relates force ( F ) ( F ) , velocity ( v ) ( v ) , mass , and resistance ( m ) ( m ) . Find R R when m = 45 , m = 45 , v = 7 , v = 7 , and F = 245. F = 245.

F = m a F = m a indicates that force ( F ) equals mass ( m ) times acceleration ( a ). Find the acceleration of a mass of 50 kg if a force of 12 N is exerted on it.

S u m = 1 1 − r S u m = 1 1 − r is the formula for an infinite series sum. If the sum is 5, find r . r .

For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question.

Solve for W : P = 2 L + 2 W P = 2 L + 2 W

Use the formula from the previous question to find the width, W , W , of a rectangle whose length is 15 and whose perimeter is 58.

Solve for f : 1 p + 1 q = 1 f f : 1 p + 1 q = 1 f

Use the formula from the previous question to find f f when p = 8 and  q = 13. p = 8 and  q = 13.

Solve for m m in the slope-intercept formula: y = m x + b y = m x + b

Use the formula from the previous question to find m m when the coordinates of the point are ( 4 , 7 ) ( 4 , 7 ) and b = 12. b = 12.

The area of a trapezoid is given by A = 1 2 h ( b 1 + b 2 ) . A = 1 2 h ( b 1 + b 2 ) . Use the formula to find the area of a trapezoid with h = 6 , b 1 = 14 , and  b 2 = 8. h = 6 , b 1 = 14 , and  b 2 = 8.

Solve for h: A = 1 2 h ( b 1 + b 2 ) A = 1 2 h ( b 1 + b 2 )

Use the formula from the previous question to find the height of a trapezoid with A = 150 , b 1 = 19 A = 150 , b 1 = 19 , and  b 2 = 11. b 2 = 11.

Find the dimensions of an American football field. The length is 200 ft more than the width, and the perimeter is 1,040 ft. Find the length and width. Use the perimeter formula P = 2 L + 2 W . P = 2 L + 2 W .

Distance equals rate times time, d = r t . d = r t . Find the distance Tom travels if he is moving at a rate of 55 mi/h for 3.5 h.

Using the formula in the previous exercise, find the distance that Susan travels if she is moving at a rate of 60 mi/h for 6.75 h.

What is the total distance that two people travel in 3 h if one of them is riding a bike at 15 mi/h and the other is walking at 3 mi/h?

If the area model for a triangle is A = 1 2 b h , A = 1 2 b h , find the area of a triangle with a height of 16 in. and a base of 11 in.

Solve for h: A = 1 2 b h A = 1 2 b h

Use the formula from the previous question to find the height to the nearest tenth of a triangle with a base of 15 and an area of 215.

The volume formula for a cylinder is V = π r 2 h . V = π r 2 h . Using the symbol π π in your answer, find the volume of a cylinder with a radius, r , r , of 4 cm and a height of 14 cm.

Solve for h: V = π r 2 h V = π r 2 h

Use the formula from the previous question to find the height of a cylinder with a radius of 8 and a volume of 16 π 16 π

Solve for r: V = π r 2 h V = π r 2 h

Use the formula from the previous question to find the radius of a cylinder with a height of 36 and a volume of 324 π . 324 π .

The formula for the circumference of a circle is C = 2 π r . C = 2 π r . Find the circumference of a circle with a diameter of 12 in. (diameter = 2 r ). Use the symbol π π in your final answer.

Solve the formula from the previous question for π . π . Notice why π π is sometimes defined as the ratio of the circumference to its diameter.

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  • Book title: College Algebra 2e
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Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Elimination Method of solving linear equations

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Graphical Method of Solving Linear Equations

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Cross Multiplication Method of solving linear equations

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

  • Linear Equations
  • Application of Linear Equations
  • Two-Variable Linear Equations
  • Linear Equations and Half Planes
  • One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Solving Linear Equations Example

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

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4.3: Solve Applications with Systems of Equations

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Learning Objectives

By the end of this section, you will be able to:

  • Solve direct translation applications
  • Solve geometry applications

Solve uniform motion applications

Before you get started, take this readiness quiz.

  • The sum of twice a number and nine is 31. Find the number. If you missed this problem, review [link] .
  • Twins Jon and Ron together earned $96,000 last year. Ron earned $8000 more than three times what Jon earned. How much did each of the twins earn? If you missed this problem, review [link] .
  • An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in four hours and the local train takes five hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains. If you missed this problem, review [link] .

Solve Direct Translation Applications

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we’ll first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

SOLVE APPLICATIONS WITH SYSTEMS OF EQUATIONS.

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose variables to represent those quantities.
  • Translate into a system of equations.
  • Solve the system of equations using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

We solved number problems with one variable earlier. Let’s see how differently it works using two variables.

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Example \(\PageIndex{2}\)

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

Example \(\PageIndex{3}\)

The sum of two numbers is \(−6\). One number is 10 less than the other. Find the numbers.

\(2, −8\)

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her \($10,000+$40\) for each training session. How many training sessions would make the salary options equal?

Example \(\PageIndex{5}\)

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

160 policies

Example \(\PageIndex{6}\)

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

As you solve each application, remember to analyze which method of solving the system of equations would be most convenient.

Example \(\PageIndex{7}\)

Translate to a system of equations and then solve:

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories for each minute of circuit training?

Example \(\PageIndex{8}\)

Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.

Example \(\PageIndex{9}\)

Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?

Erin burned 11 calories for each minute on the rowing machine and 5 calories for each minute of weight lifting.

Solve Geometry Applications

We will now solve geometry applications using systems of linear equations. We will need to add complementary angles and supplementary angles to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

COMPLEMENTARY AND SUPPLEMENTARY ANGLES

Two angles are complementary if the sum of the measures of their angles is 90 degrees.

Two angles are supplementary if the sum of the measures of their angles is 180 degrees

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

Example \(\PageIndex{10}\)

Translate to a system of equations and then solve.

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

\(\begin{array} {ll} {\textbf{Step 1. Read }\text{the problem.}} &{} \\ {\textbf{Step 2. Identify }\text{what we are looking for.}} &{\text{We are looking for the measure of each}} \\ {} &{\text{angle.}} \\ {\textbf{Step 3. Name }\text{what we are looking for.}} &{\text{Let} x=\text{ the measure of the first angle.}} \\ {} &{\hspace{3mm} y= \text{ the measure of the second angle}} \\ {\textbf{Step 4. Translate }\text{into a system of}} &{\text{The angles are complementary.}} \\ {\text{equations.}} &{\hspace{15mm} x+y=90} \\ {} &{\text{The difference of the two angles is 26}} \\ {} &{\text{degrees.}} \\ {} &{\hspace{15mm} x−y=26} \\ {} &{} \\ {} &{} \\ {\text{The system is shown.}} &{\hspace{15mm} \left\{ \begin{array} {l} x+y=90 \\ x−y=26 \end{array} \right. } \\ {} &{} \\ {} &{} \\ {\textbf{Step 5. Solve }\text{the system of equations} } &{\hspace{15mm} \left\{ \begin{array} {l} x+y=90 \\ \underline{x−y=26} \end{array} \right. } \\ {\text{by elimination.}} &{\hspace{21mm} 2x\hspace{4mm}=116} \\ {} &{\hspace{28mm} x=58} \\ {} &{} \\ {} &{} \\ {\text{Substitute }x=58\text{ into the first equation.}} &{\hspace{15mm} x+y=90} \\ {} &{\hspace{14mm} 58+y=90} \\ {} &{\hspace{22mm} y=32} \\ {\textbf{Step 6. Check }\text{the answer in the problem.}} &{} \\ {} &{} \\ {} &{} \\ {} &{} \\ {\hspace{15mm} 58+32=90\checkmark} &{} \\ {\hspace{15mm} 58−32=26\checkmark} &{} \\ {\textbf{Step 7. Answer }\text{the question.}} &{\text{The angle measures are 58 and 32 degrees.}} \end{array} \)

Example \(\PageIndex{11}\)

The difference of two complementary angles is 20 degrees. Find the measures of the angles.

The angle measures are 55 and 35.

Example \(\PageIndex{12}\)

The difference of two complementary angles is 80 degrees. Find the measures of the angles.

The angle measures are 5 and 85.

In the next example, we remember that the measures of supplementary angles add to 180.

Example \(\PageIndex{13}\)

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Example \(\PageIndex{14}\)

Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

The angle measures are 42 and 138.

Example \(\PageIndex{15}\)

Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.

The angle measures are 66 and 114.

Recall that the angles of a triangle add up to 180 degrees. A right triangle has one angle that is 90 degrees. What does that tell us about the other two angles? In the next example we will be finding the measures of the other two angles.

Example \(\PageIndex{16}\)

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Example \(\PageIndex{17}\)

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

Example \(\PageIndex{18}\)

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

Often it is helpful when solving geometry applications to draw a picture to visualize the situation.

Example \(\PageIndex{19}\)

Randall has 125 feet of fencing to enclose the part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Example \(\PageIndex{20}\)

Mario wants to put a fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

The length is 60 feet and the width is 35 feet.

Example \(\PageIndex{21}\)

Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.

The length is 60 feet and the width is 38 feet.

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was \(D=rt\) where D is the distance traveled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Example \(\PageIndex{22}\)

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

A diagram is useful in helping us visualize the situation.

.

Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, k , and Joni, j , will each drive.

.

Since \(D=r·t\) we can fill in the Distance column.

To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So,

\(\hspace{85mm} 65j=78k \nonumber \)

Also, since Kelly left later, her time will be \(\frac{1}{2}\) hour less than Joni’s time. So,

\( \hspace{105mm} k=j-\frac{1}{2} \nonumber \)

\(\begin{array} {ll} {\text{Now we have the system.}} &{\left\{ \begin{array} {l} k=j−\frac{1}{2} \\ 65j=78k \end{array} \right.} \\ {\textbf{Solve }\text{the system of equations by substitution.}} &{} \\ {} &{} \\ {\text{Substitute }k=j−12\text{ into the second equation,}} &{} \\ {\text{then solve for }j.} &{} \\ {} &{65j=78k} \\ {} &{65j=78(j−\frac{1}{2})} \\ {} &{65j=78j−39} \\ {} &{−13j=−39} \\ {} &{j=3} \\{\begin{array} {l} {\text{To find Kelly’s time, substitute }j=3 \text{ into the first}} \\ {\text{equation, then solve for }k.} \end{array} } &{k=j−\frac{1}{2}} \\ {} &{k=3−\frac{1}{2} } \\ {} &{k=\frac{5}{2} \text{ or } k=2\frac{1}{2}} \\ {\textbf{Check }\text{the answer in the problem.}} &{} \\ {\begin{array} {lllll} {\text{Joni}} &{3 \text{ hours}} &{(65\text{ mph})} &= &{195\text{ miles}} \\ {\text{Kelly}} &{2\frac{1}{2} \text{ hours}} &{(78\text{ mph})} &= &{195\text{ miles}} \end{array}} &{} \\ {\text{Yes, they will have traveled the same distance}} &{} \\{\text{when they meet.}} &{} \\ {\textbf{Answer }\text{the question.}} &{} \\ {} &{\text{Kelly will catch up to Joni in}} \\ {} &{2\frac{1}{2}\text{ hours. By then, Joni will}} \\ {} &{\text{have traveled }3 \text{ hours.}} \\ \end{array}\)

Example \(\PageIndex{23}\)

Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

It will take Clark 4 hours to catch Mitchell.

Example \(\PageIndex{24}\)

Charlie left his mother’s house traveling at an average speed of 36 miles per hour. His sister Sally left 15 minutes \((\frac{1}{4} \text{ hour})\) later traveling the same route at an average speed of 42 miles per hour. How long before Sally catches up to Charlie?

It will take Sally \(112\) hours to catch up to Charlie.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c .

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is \(b+c\).

Figure shows a boat and two horizontal arrows, both pointing left. The one to the left of the boat is b and the one to the right is c.

Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is \(b−c\).

Figure shows a boat and two horizontal arrows to its left. One, labeled b, points left and the other, labeled c, points right.

We’ll put some numbers to this situation in the next example.

Example \(\PageIndex{25}\)

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Example \(\PageIndex{26}\)

A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

The rate of the boat is 11 mph and the rate of the current is 1 mph.

Example \(\PageIndex{27}\)

Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

The speed of the canoe is 7 mph and the speed of the current is 1 mph.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the next example. A wind current in the same direction as the plane is flying is called a tailwind . A wind current blowing against the direction of the plane is called a headwind .

Example \(\PageIndex{28}\)

A private jet can fly 1,095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Example \(\PageIndex{29}\)

A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1,035 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 235 mph and the speed of the wind is 30 mph.

Example \(\PageIndex{30}\)

A commercial jet can fly 1,728 miles in 4 hours with a tailwind but only 1,536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 408 mph and the speed of the wind is 24 mph.

Access this online resource for additional instruction and practice with systems of equations.

  • Systems of Equations

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Heavy Machinery Meets AI

  • Vijay Govindarajan
  • Venkat Venkatraman

more problem solving using equations

Until recently most incumbent industrial companies didn’t use highly advanced software in their products. But now the sector’s leaders have begun applying generative AI and machine learning to all kinds of data—including text, 3D images, video, and sound—to create complex, innovative designs and solve customer problems with unprecedented speed.

Success involves much more than installing computers in products, however. It requires fusion strategies, which join what manufacturers do best—creating physical products—with what digital firms do best: mining giant data sets for critical insights. There are four kinds of fusion strategies: Fusion products, like smart glass, are designed from scratch to collect and leverage information on product use in real time. Fusion services, like Rolls-Royce’s service for increasing the fuel efficiency of aircraft, deliver immediate customized recommendations from AI. Fusion systems, like Honeywell’s for building management, integrate machines from multiple suppliers in ways that enhance them all. And fusion solutions, such as Deere’s for increasing yields for farmers, combine products, services, and systems with partner companies’ innovations in ways that greatly improve customers’ performance.

Combining digital and analog machines will upend industrial companies.

Idea in Brief

The problem.

Until recently most incumbent industrial companies didn’t use the most advanced software in their products. But competitors that can extract complex designs, insights, and trends using generative AI have emerged to challenge them.

The Solution

Industrial companies must develop strategies that fuse what they do best—creating physical products—with what digital companies do best: using data and AI to parse enormous, interconnected data sets and develop innovative insights.

The Changes Required

Companies will have to reimagine analog products and services as digitally enabled offerings, learn to create new value from data generated by the combination of physical and digital assets, and partner with other companies to create ecosystems with an unwavering focus on helping customers solve problems.

For more than 187 years, Deere & Company has simplified farmwork. From the advent of the first self-scouring plow, in 1837, to the launch of its first fully self-driving tractor, in 2022, the company has built advanced industrial technology. The See & Spray is an excellent contemporary example. The automated weed killer features a self-propelled, 120-foot carbon-fiber boom lined with 36 cameras capable of scanning 2,100 square feet per second. Powered by 10 onboard vision-processing units handling almost four gigabytes of data per second, the system uses AI and deep learning to distinguish crops from weeds. Once a weed is identified, a command is sent to spray and kill it. The machine moves through a field at 12 miles per hour without stopping. Manual labor would be more expensive, more time-consuming, and less reliable than the See & Spray. By fusing computer hardware and software with industrial machinery, it has helped farmers decrease their use of herbicide by more than two-thirds and exponentially increase productivity.

  • Vijay Govindarajan is the Coxe Distinguished Professor at Dartmouth College’s Tuck School of Business, an executive fellow at Harvard Business School, and faculty partner at the Silicon Valley incubator Mach 49. He is a New York Times and Wall Street Journal bestselling author. His latest book is Fusion Strategy: How Real-Time Data and AI Will Power the Industrial Future . His Harvard Business Review articles “ Engineering Reverse Innovations ” and “ Stop the Innovation Wars ” won McKinsey Awards for best article published in HBR. His HBR articles “ How GE Is Disrupting Itself ” and “ The CEO’s Role in Business Model Reinvention ” are HBR all-time top-50 bestsellers. Follow him on LinkedIn . vgovindarajan
  • Venkat Venkatraman is the David J. McGrath Professor at Boston University’s Questrom School of Business, where he is a member of both the information systems and strategy and innovation departments. His current research focuses on how companies develop winning digital strategies. His latest book is Fusion Strategy: How Real-Time Data and AI Will Power the Industrial Future.  Follow him on LinkedIn . NVenkatraman

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VIDEO

  1. PROBLEM-SOLVING USING WORKING BACKWARDS STRATEGY

  2. Equations--More Problem Solving

  3. how to solve this equation?

  4. Math 10C Unit 6 Trigonometry L7: More Problem Solving Using Trigonometric Ratios

  5. problem solving using determinant #matrices #maths

  6. Some more problem solving techniques

COMMENTS

  1. Solving Equations

    In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value. Combine like terms. Simplify the equation by using the opposite operation to both sides. Isolate the variable on one side of the equation. Solving equations examples

  2. Solving Equations

    Solving Equations What is an Equation? An equation says that two things are equal. It will have an equals sign "=" like this: x − 2 = 4 That equations says: what is on the left (x − 2) equals what is on the right (4) So an equation is like a statement " this equals that " What is a Solution?

  3. Solving equations & inequalities

    There are lots of strategies we can use to solve equations. Let's explore some different ways to solve equations and inequalities. We'll also see what it takes for an equation to have no solution, or infinite solutions. Linear equations with variables on both sides Learn Why we do the same thing to both sides: Variable on both sides

  4. Equation Solver

    To solve your equation using the Equation Solver, type in your equation like x+4=5. The solver will then show you the steps to help you learn how to solve it on your own. Solving Equations Video Lessons

  5. Microsoft Math Solver

    Solve trigonometry Get step-by-step explanations See how to solve problems and show your work—plus get definitions for mathematical concepts Graph your math problems Instantly graph any equation to visualize your function and understand the relationship between variables Practice, practice, practice

  6. 2.5: Applications of Linear Equations

    Guidelines for Setting Up and Solving Word Problems. Step 1: Read the problem several times, identify the key words and phrases, and organize the given information.; Step 2: Identify the variables by assigning a letter or expression to the unknown quantities.; Step 3: Translate and set up an algebraic equation that models the problem.; Step 4: Solve the resulting algebraic equation.

  7. Algebraic word problems

    Solving algebraic word problems requires us to combine our ability to create equations and solve them. To solve an algebraic word problem: Define a variable. Write an equation using the variable. Solve the equation. If the variable is not the answer to the word problem, use the variable to calculate the answer.

  8. Problem solving with equations

    Task 1. The father is x years old. Mom is two years younger. The son is three times younger than the father. Write the age of each using expressions. Solution: Task 2. Father is x years old, mother is 2 years younger than father. Son is 3 times younger than father, daughter is 3 times younger than mother. Write the age of each using expressions.

  9. 5.4: Solve Applications with Systems of Equations

    Simplify and add the equations. Solve for c. Substitute c = 8.3 into one of the original equations to solve for e. Step 6. Check the answer in the problem. Check the math on your own. Step 7. Answer the question. Jenna burns 8.3 calories per minute circuit training and 11.2 calories per minute while on the elliptical trainer.

  10. Solving basic equations & inequalities (one variable, linear)

    Algebra (all content) 20 units · 412 skills. Unit 1 Introduction to algebra. Unit 2 Solving basic equations & inequalities (one variable, linear) Unit 3 Linear equations, functions, & graphs. Unit 4 Sequences. Unit 5 System of equations. Unit 6 Two-variable inequalities. Unit 7 Functions. Unit 8 Absolute value equations, functions, & inequalities.

  11. 4.9: Strategies for Solving Applications and Equations

    Number word problems give some clues about one or more numbers and we use these clues to write an equation. Number word problems provide good practice for using the Problem Solving Strategy. EXAMPLE \(\PageIndex{4}\)

  12. 5.7 Use a Problem-Solving Strategy

    Choose a variable to represent the number. Let the number. Step 4. Translate. Restate the problem as one sentence. Translate into an equation. Step 5. Solve the equation. Subtract 7 from each side and simplify.

  13. Solve

    QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and ...

  14. Algebra

    Here are a set of practice problems for the Solving Equations and Inequalities chapter of the Algebra notes. If you'd like a pdf document containing the solutions the download tab above contains links to pdf's containing the solutions for the full book, chapter and section. At this time, I do not offer pdf's for solutions to individual problems.

  15. Problem Solving

    W = Width In the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution. This video shows a similar garden box problem.

  16. Equation Solver

    Step 1: Enter the Equation you want to solve into the editor. The equation calculator allows you to take a simple or complex equation and solve by best method possible. Step 2: Click the blue arrow to submit and see the result! The equation solver allows you to enter your problem and solve the equation to see the result.

  17. Algebra

    Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. Paul's Online Notes ... 4.9 More Optimization Problems; 4.10 L'Hospital's Rule and Indeterminate Forms; 4.11 Linear Approximations; 4.12 Differentials; 4 ...

  18. 2.3 Models and Applications

    Given a real-world problem, model a linear equation to fit it. Identify known quantities. Assign a variable to represent the unknown quantity. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first. Write an equation interpreting the words as mathematical operations. Solve the equation.

  19. Elimination method review (systems of linear equations)

    Example 1. We're asked to solve this system of equations: 2 y + 7 x = − 5 5 y − 7 x = 12. We notice that the first equation has a 7 x term and the second equation has a − 7 x term. These terms will cancel if we add the equations together—that is, we'll eliminate the x terms: 2 y + 7 x = − 5 + 5 y − 7 x = 12 7 y + 0 = 7. Solving for ...

  20. Solving Linear Equations

    Solving Linear Equations. Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation.

  21. Step-by-Step Calculator

    Equations Inequalities System of Equations System of Inequalities ... To solve math problems step-by-step start by reading the problem carefully and understand what you are being asked to find. Next, identify the relevant information, define the variables, and plan a strategy for solving the problem. Show more; en. Related Symbolab blog posts ...

  22. Equation Calculator

    Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. ... Get full access to all Solution Steps for any math problem Continue with Google By continuing, ... get the variable on one side of the equation by using inverse operations. Show more; Why users love our ...

  23. Solving Differential Equations With Neural Networks

    Its only purpose is to solve a differential equation within the time domain it was crafted to solve. Hence, to test it, it's only fair that we use the time domain it was trained on. Fig. 3 shows a comparison between the NN prediction and the theoretical answer (that is, the analytical solution).

  24. 10 Best strategies for solving math word problems in 2024

    6. Use Estimation to Predict Answers. Estimation is a valuable skill in solving math word problems, as it allows students to predict the answer's ballpark figure before solving it precisely. Teaching students to use estimation can help them check their answers for reasonableness and avoid common mistakes.

  25. Find the AI Approach That Fits the Problem You're Trying to Solve

    Summary. AI moves quickly, but organizations change much more slowly. What works in a lab may be wrong for your company right now. If you know the right questions to ask, you can make better ...

  26. 4.3: Solve Applications with Systems of Equations

    Step 5. Solve the system of equations. We will use substitution since the first equation is solved for a. Substitute 3b+103b+10 for a in the second equation. Solve for b. Substitute b=20b=20 into the first equation and then solve for a. Step 6. Check the answer in the problem. We will leave this to you! Step 7. Answer the question.

  27. Heavy Machinery Meets AI

    The automated weed killer features a self-propelled, 120-foot carbon-fiber boom lined with 36 cameras capable of scanning 2,100 square feet per second. Powered by 10 onboard vision-processing ...