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Solving Systems of Equations Real World Problems

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method . In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Now we are ready to apply these strategies to solve real world problems! Are you ready? First let's look at some guidelines for solving real world problems and then we'll look at a few examples.

Steps For Solving Real World Problems

  • Highlight the important information in the problem that will help write two equations.
  • Define your variables
  • Write two equations
  • Use one of the methods for solving systems of equations to solve.
  • Check your answers by substituting your ordered pair into the original equations.
  • Answer the questions in the real world problems. Always write your answer in complete sentences!

Ok... let's look at a few examples. Follow along with me. (Having a calculator will make it easier for you to follow along.)

Example 1: Systems Word Problems

You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs sold and the number of sodas sold. How many hot dogs were sold and how many sodas were sold?

1.  Let's start by identifying the important information:

  • hot dogs cost $1.50
  • Sodas cost $0.50
  • Made a total of $78.50
  • Sold 87 hot dogs and sodas combined

2.  Define your variables.

  • Ask yourself, "What am I trying to solve for? What don't I know?

In this problem, I don't know how many hot dogs or sodas were sold. So this is what each variable will stand for. (Usually the question at the end will give you this information).

Let x = the number of hot dogs sold

Let y = the number of sodas sold

3. Write two equations.

One equation will be related to the price and one equation will be related to the quantity (or number) of hot dogs and sodas sold.

1.50x + 0.50y = 78.50    (Equation related to cost)

 x + y = 87   (Equation related to the number sold)

4.  Solve! 

We can choose any method that we like to solve the system of equations. I am going to choose the substitution method since I can easily solve the 2nd equation for y.

Solving a systems using substitution

5. Think about what this solution means.

x is the number of hot dogs and x = 35. That means that 35 hot dogs were sold.

y is the number of sodas and y = 52. That means that 52 sodas were sold.

6.  Write your answer in a complete sentence.

35 hot dogs were sold and 52 sodas were sold.

7.  Check your work by substituting.

1.50x + 0.50y = 78.50

1.50(35) + 0.50(52) = 78.50

52.50 + 26 = 78.50

35 + 52 = 87

Since both equations check properly, we know that our answers are correct!

That wasn't too bad, was it? The hardest part is writing the equations. From there you already know the strategies for solving. Think carefully about what's happening in the problem when trying to write the two equations.

Example 2: Another Word Problem

You and a friend go to Tacos Galore for lunch. You order three soft tacos and three burritos and your total bill is $11.25. Your friend's bill is $10.00 for four soft tacos and two burritos. How much do soft tacos cost? How much do burritos cost?

  • 3 soft tacos + 3 burritos cost $11.25
  • 4 soft tacos + 2 burritos cost $10.00

In this problem, I don't know the price of the soft tacos or the price of the burritos.

Let x = the price of 1 soft taco

Let y = the price of 1 burrito

One equation will be related your lunch and one equation will be related to your friend's lunch.

3x + 3y = 11.25  (Equation representing your lunch)

4x + 2y = 10   (Equation representing your friend's lunch)

We can choose any method that we like to solve the system of equations. I am going to choose the combinations method.

Solving Systems Using Combinations

5. Think about what the solution means in context of the problem.

x = the price of 1 soft taco and x = 1.25.

That means that 1 soft tacos costs $1.25.

y = the price of 1 burrito and y = 2.5.

That means that 1 burrito costs $2.50.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

If you have difficulty with real world problems, you can find more examples and practice problems in the Algebra Class E-course.

Take a look at the questions that other students have submitted:

solving linear systems by graphing word problems

Problem about the WNBA

Systems problem about ages

Problem about milk consumption in the U.S.

Vans and Buses? How many rode in each?

Telephone Plans problem

Systems problem about hats and scarves

Apples and guavas please!

How much did Alice spend on shoes?

All about stamps

Going to the movies

Small pitchers and large pitchers - how much will they hold?

Chickens and dogs in the farm yard

  • System of Equations
  • Systems Word Problems

solving linear systems by graphing word problems

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Systems of Linear Equations and Word Problems

Note that we saw how to solve linear inequalities here in the Coordinate System and Graphing Lines section . Note also that we solve Algebra Word Problems without Systems here , and we solve systems using matrices in the Matrices and Solving Systems with Matrices  section here.

Introduction to Systems

“Systems of equations” just means that we are dealing with more than one equation and variable. So far, we’ve basically just played around with the equation for a line, which is $ y=mx+b$. Let’s say we have the following situation:

Now, you can always do “guess and check” to see what would work, but you might as well use algebra! It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.

The first trick in problems like this is to figure out what we want to know. This will help us decide what variables (unknowns) to use. What we want to know is how many pairs of jeans we want to buy (let’s say “$ j$”) and how many dresses we want to buy (let’s say “$ d$”). Always write down what your variables will be:

Let $ j=$ the number of jeans you will buy Let $ d=$ the number of dresses you’ll buy

Like we did before, let’s translate word-for-word from math to English:

Now we have the 2 equations as shown below. Notice that the $ j$ variable is just like the $ x$ variable and the $ d$ variable is just like the $ y$. It’s easier to put in $ j$   and $ d$ so we can remember what they stand for when we get the answers.

This is what we call a system, since we have to solve for more than one variable – we have to solve for 2 here. The cool thing is to solve for 2   variables , you typically need 2   equations , to solve for 3   variables , you need 3   equations , and so on. That’s easy to remember, right?

We need to get an answer that works in both equations ; this is what we’re doing when we’re solving; this is called solving simultaneous systems , or solving system simultaneously . There are several ways to solve systems; we’ll talk about graphing first.

Solving Systems by Graphing

Remember that when you graph a line, you see all the different coordinates (or $ x/y$ combinations) that make the equation work. In systems, you have to make both equations work, so the intersection of the two lines shows the point that fits both equations (assuming the lines do in fact intersect; we’ll talk about that later).  The points of intersections satisfy both equations simultaneously. 

Put these equations into the $ y=mx+b$ ($ d=mj+b$) format, by solving for the $ d$ (which is like the $ y$):

$ \displaystyle j+d=6;\text{ }\,\text{ }\text{solve for }d:\text{ }d=-j+6\text{ }$

$ \displaystyle 25j+50d=200;\text{ }\,\,\text{solve for }d:\text{ }d=\frac{{200-25j}}{{50}}=-\frac{1}{2}j+4$

Now graph both lines:

Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the  Exponents and Radicals in Algebra section. Also, t here are some examples of systems of inequality  here in the Coordinate System and Graphing Lines section .

Solving Systems with Substitution

Substitution is the favorite way to solve for many students! It involves exactly what it says: substituting one variable in another equation so that you only have one variable in that equation. Here is the same problem:

You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50 .  You really, really want to take home 6 items of clothing because you “need” that many new things. How many pairs of jeans and how many dresses you can buy so you use the whole $200 (tax not included)?

Below are our two equations, and let’s solve for “$ d$” in terms of “$ j$” in the first equation. Then, let’s substitute what we got for “$ d$” into the next equation. Even though it doesn’t matter which equation you start with, remember to always pick the “easiest” equation first (one that we can easily solve for a variable) to get a variable by itself.

We could buy 4 pairs of jeans and 2 dresses . Note that we could have also solved for “$ j$” first; it really doesn’t matter. You’ll want to pick the variable that’s most easily solved for. Let’s try another substitution problem that’s a little bit different:

Solving Systems with Linear Combination or Elimination

Probably the most useful way to solve systems is using linear combination, or linear elimination. The reason it’s most useful is that usually in real life we don’t have one variable in terms of another (in other words, a “$ y=$” situation).

The main purpose of the linear combination method is to add or subtract the equations so that one variable is eliminated. We can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below. We are using the Additive Property of Equality , Subtraction Property of Equality , Multiplicative Property of Equality , and/or Division Property of Equality that we saw here in the Types of Numbers and Algebraic Properties section :

If we have a set of 2 equations with 2 unknowns, for example, we can manipulate them by adding, multiplying or subtracting (we usually prefer adding) so that we get one equation with one variable. Let’s use our previous problem:

We could buy 4 pairs of jeans and 2 dresses .

Here’s another example:

Types of equations

In the example above, we found one unique solution to the set of equations. Sometimes, however, for a set of equations, there are no solutions (when lines are parallel) or an infinite number of solutions or infinitely many solutions (when the two lines are actually the same line, and one is just a “multiple” of the other).

When there is at least one solution , the equations are consistent equations , since they have a solution. When there is only one solution, the system is called independent , since they cross at only one point. When equations have infinite solutions, they are the same equation, are consistent , and are called dependent or coincident (think of one just sitting on top of the other).

When equations have no solutions , they are called inconsistent equations , since we can never get a solution . 

Here are graphs of inconsistent and dependent equations that were created on a graphing calculator:

Systems with Three Equations

Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns to solve for.

Let’s say at the same store, they also had pairs of shoes for $20 and we managed to get $60 more to spend! Now we have a new problem. To spend the even $260 , how many pairs of jeans, dresses, and pairs of shoes should we get if want, for example, exactly 10 total items (Remember that jeans cost $25 each and dresses cost $50 each).

Let’s let $ j=$ the number of pair of jeans, $ d=$ the number of dresses, and $ s=$ the number of pairs of shoes we should buy. So far, we’ll have the following equations:

$ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+\,20s=260\end{array}$

We’ll need another equation, since for three variables, we need three equations (otherwise, we theoretically might have infinite ways to solve the problem). In this type of problem, you would also need something like this: We want twice as many pairs of jeans as pairs of shoes . Now, since we have the same number of equations as variables , we can potentially get one solution for the system of equations. Here are the three equations:

We’ll learn later how to put these in our calculator to easily solve using matrices (see the  Matrices and Solving Systems with Matrices section). For now, we can use two equations at a time to eliminate a variable (using substitution and/or elimination), and keep doing this until we’ve solved for all variables. These can get really difficult to solve, but remember that in “real life”, there are computers to do all this work!

Remember again, that if we ever get to a point where we end up with something like this, it means there are an infinite number of solutions : $ 4=4$  (variables are gone and a number equals another number and they are the same). And if we up with something like this, it means there are no solutions : $ 5=2$ (variables are gone and two numbers are left and they don’t equal each other).

Let’s solve our system:      $ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+20s=260\\j=2s\end{array}$ :

We could buy 6 pairs of jeans, 1 dress, and 3 pairs of shoes .

Here’s one more example of a three-variable system of equations, where we’ll only use linear elimination:

$ \displaystyle \begin{align}5x-6y-\,7z\,&=\,7\\6x-4y+10z&=\,-34\\2x+4y-\,3z\,&=\,29\end{align}$

I know – this is really difficult stuff! But if you do it step-by-step and keep using the equations you need with the right variables, you can do it. Think of it like a puzzle – you may not know exactly where you’re going, but do what you can in baby steps, and you’ll get there (sort of like life sometimes, right?!). And we’ll learn much easier ways to do these types of problems.

Algebra Word Problems with Systems

Let’s do more word problems; you’ll notice that many of these are the same type that we did earlier in the Algebra Word Problems  section , but now we can use more than one variable. This will actually make the problems easier! Again, when doing these word problems:

  • If you’re wondering what the variables (or unknowns) should be when working on a word problem, look at what the problem is asking. These are usually (but not always) what your variables are!
  • If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!

Here are some problems:

Investment Word Problem

We also could have set up this problem with a table:

Mixture Word Problems

Here’s a mixture word problem . With mixture problems, remember if the problem calls for a pure solution or concentrate , use 100% (if the percentage is that solution) or 0% (if the percentage is another solution).

Let’s do the math (use substitution )!

$ \displaystyle \begin{array}{c}x\,\,+\,\,y=10\\.01x+.035y=10(.02)\end{array}$          $ \displaystyle \begin{array}{c}\,y=10-x\\.01x+.035(10-x)=.2\\.01x\,+\,.35\,\,-\,.035x=.2\\\,-.025x=-.15;\,\,\,\,\,x=6\\\,y=10-6=4\end{array}$

We would need 6 liters of the 1% milk, and 4 liters of the 3.5% milk.

Here’s another mixture problem:

$ \displaystyle \begin{array}{c}x+y=50\\8x+4y=50\left( {6.4} \right)\end{array}$                   $ \displaystyle \begin{array}{c}y=50-x\\8x+4\left( {50-x} \right)=320\\8x+200-4x=320\\4x=120\,;\,\,\,\,x=30\\y=50-30=20\\8x+4y=50(6.4)\end{array}$

We would need 30 pounds of the $8 coffee bean, and 20 pounds of the $4 coffee bean. See how similar this problem is to the one where we use percentages?

Distance Word Problem:

Here’s a distance word problem using systems ; distance problems have to do with an object’s speed, time, and distance. Note that, as well as the distance word problem here in the Algebra Word Problems section , there’s an example of a Parametric Distance Problem here in the Parametric Equations section .

Which Plumber Problem

Many word problems you’ll have to solve have to do with an initial charge or setup charge, and a charge or rate per time period. In these cases, the initial charge will be the $ \boldsymbol {y}$ -intercept , and the  rate will be the slope . Here is an example:

Geometry Word Problem:

Many times, we’ll have a geometry problem as an algebra word problem; these might involve perimeter, area, or sometimes angle measurements (so don’t forget these things!). Let’s do one involving angle measurements.

See – these are getting easier! Here’s one that’s a little tricky though:

Work Problem : 

Let’s do a “ work problem ” that is typically seen when studying Rational Equations (fraction with variables in them) and can be found here in the Rational Functions, E quations and Inequalities  section .

Note that there’s also a simpler version of this problem here in the Direct, Inverse, Joint and Combined Variation section .

Three Variable Word Problem:

Let’s do one more with three equations and three unknowns:

The “Candy” Problem

Sometimes we get lucky and can solve a system of equations where we have more unknowns (variables) then equations. (Actually, I think it’s not so much luck, but having good problem writers!) Here’s one like that:

There are more Systems Word Problems in the  Matrices and Solving Systems with Matrices section , Linear Programming section , and Right Triangle Trigonometry section .

Understand these problems, and practice, practice, practice!

For Practice : Use the Mathway  widget below to try a  Systems of Equations  problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve by Substitution or Solve by Addition/Elimination  to see the answer .

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps , or Click Here , you can register at Mathway for a free trial , and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Algebraic Functions, including Domain and Range   – you’re ready! 

solving linear systems by graphing word problems

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System-of-Equations Word Problems

Exercises More Exercises

Many problems lend themselves to being solved with systems of linear equations. In "real life", these problems can be incredibly complex. This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics.

In your studies, however, you will generally be faced with much simpler problems. What follows are some typical examples.

The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?

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System of Equations Word Problems

In the past , I would have set this up by picking a variable for one of the groups (say, " c " for "children") and then use "(total) less (what I've already accounted for)" (in this case, " 2200 –  c ") for the other group. Using a system of equations, however, allows me to use two different variables for the two different unknowns.

number of adults: a

number of children: c

With these variables, I can create equations for the totals they've given me:

total number: a + c = 2200

total income: 4 a + 1.5 c = 5050

Now I can solve the system for the number of adults and the number of children. I will solve the first equation for one of the variables, and then substitute the result into the other equation:

a = 2200 – c

4(2200 – c ) + 1.5 c = 5050

8800 – 4 c + 1.5 c = 5050

8800 – 2.5 c = 5050

–2.5 c = –3750

Now I can back-solve for the value of the other variable:

a = 2200 – (1500) = 700

I have values for my two variables. I can look back at my definitions for the variables to interpret these values. To answer the original question, there were:

1500 children and 700 adults.

You will probably start out with problems which, like the one above, seem very familiar. But you will then move on to more complicated problems.

The sum of the digits of a two-digit number is 7 . When the digits are reversed, the number is increased by 27 . Find the number.

The trick here is to work with the digits explicitly. I'll use " t " for the "tens" digit of the original number and " u " for the "units" (or "ones") digit. I then have:

The ten's digit stands for "ten times of this digit's value". Just as "26" is "10 times 2, plus 6 times 1", so also the two-digit number they've given me will be ten times the "tens" digit, plus one times the "units" digit. In other words:

original number: 10 t + 1 u

The new number has the values of the digits (represented by the variables) in reverse order. This gives me:

new number: 10 u + 1 t

And this new number is twenty-seven more than the original number. The keyword "is" means "equals", so I get:

(new number) is (old number) increased by (twenty-seven)

10 u + 1 t = (10 t + 1 u ) + 27

Now I have a system of equations that I can solve:

10 u + t = 10 t + u + 27

First I'll simplify the second equation:

9 u – 9 t = 27

u – t = 3

After reordering the variables in the first equation, I now have:

Adding down , I get:

Then t = 2 . Back-solving, this means that the original number was 25 and the new number (gotten by switching the digits) is 52 . Since 52 – 25 = 27 , this solution checks out.

The number is 25 .

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Find the equation of the parabola that passes through the points (–1, 9) , (1, 5) , and (2, 12) .

Recalling that a parabola has a quadratic as its equation, I know that I am looking for an equation of the form ax 2  +  bx  +  c = y . Also, I know that points are of the form ( x ,  y ) . Practically speaking, this mean that, in each of these points, they have given me values for x and y that make the quadratic equation true. Plugging the three points in the general equation for a quadratic, I get a system of three equations, where the variables stand for the unknown coefficients of that quadratic:

a (–1) 2 + b (–1) + c = 9

a (1) 2 + b (1) + c = 5

a (2) 2 + b (2) + c = 12

Simplifying the three equations, I get:

1 a – b + c = 9

1 a + b + c = 5

4 a + 2 b + c = 12

I won't display the solving of this problem, but the result is that a  = 3, b  = –2, and c  = 4 , so the equation they're wanting is:

y = 3 x 2 – 2 x + 4

You may also see similar exercises referring to circles, using:

x 2 + y 2 + bx + cy + d = 0

...or other conics, though parabolas are the most common. Keep in mind that projectile problems (like shooting an arrow up in the air or dropping a penny from the roof of a tall building) are also parabola problems, using:

–( 1 / 2 ) gt 2 + v 0 t + h 0 = s

...where h 0 is the original height, v 0 is the initial velocity, s is the height at time t , usually measured in seconds, and g refers to gravity, being 9.8 if you're working in meters and 32 if you're working in feet).

All of these different permutations of the above example work the same way: Take the general equation for the curve, plug in the given points, and solve the resulting system of equations for the values of the coefficients. Warning: If you see an exercise of this sort in the homework, be advised that you may be expected to know the forms of the general equations (such as " ax 2 + bx + c = y " for parabolas) on the next text.

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solving linear systems by graphing word problems

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How to solve systems of equations by Graphing

Step by step tutorial for systems of linear equations (in 2 variables)

Video on Solving by Graphing

The Graph Method

What is the solution of the following system of equations?

$$ y = x + 1 \\ y = 2x $$

Graph both equations

picture of systems of equations

On the right, the graph of the two lines

The solution of the system is the point of intersection : (1, 2)

picture of systems of equations

Practice Problems

Use the graph method to solve the system of equations below

$$ y = 2x +1 \\ y = 4x -1 $$

System linear equations answer

The solution of this system is the point of intersection : (1,3).

System linear equations answer

Solve the following system of linear equations by graphing .

$$ \text{ A) } 2y = 4x + 2 \\ \text{ B) }2y = -x + 7 $$

Rewrite each equation in slope intercept form

$$ \text{ A) } 2y = 4x + 2 \\ \frac{1}{2} 2y = \frac{1}{2}(4x+2) \\ y = 2x +1 $$

$$ \text{ B) } 2y = 8x - 2 \\ \frac{1}{2} 2y = \frac{1}{2}( 8x - 2) \\ y = 4x +1 $$

This system of lines is the same system that we looked at in the last example.

Graph each equation to find the point of intersection --which is the solution. (same as earlier problem)

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  • 4.1 Solve Systems of Linear Equations with Two Variables
  • Introduction
  • 1.1 Use the Language of Algebra
  • 1.2 Integers
  • 1.3 Fractions
  • 1.4 Decimals
  • 1.5 Properties of Real Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Use a General Strategy to Solve Linear Equations
  • 2.2 Use a Problem Solving Strategy
  • 2.3 Solve a Formula for a Specific Variable
  • 2.4 Solve Mixture and Uniform Motion Applications
  • 2.5 Solve Linear Inequalities
  • 2.6 Solve Compound Inequalities
  • 2.7 Solve Absolute Value Inequalities
  • 3.1 Graph Linear Equations in Two Variables
  • 3.2 Slope of a Line
  • 3.3 Find the Equation of a Line
  • 3.4 Graph Linear Inequalities in Two Variables
  • 3.5 Relations and Functions
  • 3.6 Graphs of Functions
  • 4.2 Solve Applications with Systems of Equations
  • 4.3 Solve Mixture Applications with Systems of Equations
  • 4.4 Solve Systems of Equations with Three Variables
  • 4.5 Solve Systems of Equations Using Matrices
  • 4.6 Solve Systems of Equations Using Determinants
  • 4.7 Graphing Systems of Linear Inequalities
  • 5.1 Add and Subtract Polynomials
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  • 6.1 Greatest Common Factor and Factor by Grouping
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  • 6.3 Factor Special Products
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  • 6.5 Polynomial Equations
  • 7.1 Multiply and Divide Rational Expressions
  • 7.2 Add and Subtract Rational Expressions
  • 7.3 Simplify Complex Rational Expressions
  • 7.4 Solve Rational Equations
  • 7.5 Solve Applications with Rational Equations
  • 7.6 Solve Rational Inequalities
  • 8.1 Simplify Expressions with Roots
  • 8.2 Simplify Radical Expressions
  • 8.3 Simplify Rational Exponents
  • 8.4 Add, Subtract, and Multiply Radical Expressions
  • 8.5 Divide Radical Expressions
  • 8.6 Solve Radical Equations
  • 8.7 Use Radicals in Functions
  • 8.8 Use the Complex Number System
  • 9.1 Solve Quadratic Equations Using the Square Root Property
  • 9.2 Solve Quadratic Equations by Completing the Square
  • 9.3 Solve Quadratic Equations Using the Quadratic Formula
  • 9.4 Solve Equations in Quadratic Form
  • 9.5 Solve Applications of Quadratic Equations
  • 9.6 Graph Quadratic Functions Using Properties
  • 9.7 Graph Quadratic Functions Using Transformations
  • 9.8 Solve Quadratic Inequalities
  • 10.1 Finding Composite and Inverse Functions
  • 10.2 Evaluate and Graph Exponential Functions
  • 10.3 Evaluate and Graph Logarithmic Functions
  • 10.4 Use the Properties of Logarithms
  • 10.5 Solve Exponential and Logarithmic Equations
  • 11.1 Distance and Midpoint Formulas; Circles
  • 11.2 Parabolas
  • 11.3 Ellipses
  • 11.4 Hyperbolas
  • 11.5 Solve Systems of Nonlinear Equations
  • 12.1 Sequences
  • 12.2 Arithmetic Sequences
  • 12.3 Geometric Sequences and Series
  • 12.4 Binomial Theorem

Learning Objectives

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

  • Determine whether an ordered pair is a solution of a system of equations
  • Solve a system of linear equations by graphing
  • Solve a system of equations by substitution
  • Solve a system of equations by elimination
  • Choose the most convenient method to solve a system of linear equations

Be Prepared 4.1

Before you get started, take this readiness quiz.

For the equation y = 2 3 x − 4 , y = 2 3 x − 4 , ⓐ Is ( 6 , 0 ) ( 6 , 0 ) a solution? ⓑ Is ( −3 , −2 ) ( −3 , −2 ) a solution? If you missed this problem, review Example 3.2 .

Be Prepared 4.2

Find the slope and y -intercept of the line 3 x − y = 12 . 3 x − y = 12 . If you missed this problem, review Example 3.16 .

Be Prepared 4.3

Find the x- and y -intercepts of the line 2 x − 3 y = 12 . 2 x − 3 y = 12 . If you missed this problem, review Example 3.8 .

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations .

System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations .

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

A linear equation in two variables, such as 2 x + y = 7 , 2 x + y = 7 , has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) ( x , y ) that make both equations true. These are called the solutions of a system of equations .

Solutions of a System of Equations

The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair ( x , y ) . ( x , y ) .

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Example 4.1

Determine whether the ordered pair is a solution to the system { x − y = −1 2 x − y = −5 . { x − y = −1 2 x − y = −5 .

ⓐ ( −2 , −1 ) ( −2 , −1 ) ⓑ ( −4 , −3 ) ( −4 , −3 )

Determine whether the ordered pair is a solution to the system { 3 x + y = 0 x + 2 y = −5 . { 3 x + y = 0 x + 2 y = −5 .

ⓐ ( 1 , −3 ) ( 1 , −3 ) ⓑ ( 0 , 0 ) ( 0 , 0 )

Determine whether the ordered pair is a solution to the system { x − 3 y = −8 − 3 x − y = 4 . { x − 3 y = −8 − 3 x − y = 4 .

ⓐ ( 2 , −2 ) ( 2 , −2 ) ⓑ ( −2 , 2 ) ( −2 , 2 )

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

Example 4.2

How to solve a system of equations by graphing.

Solve the system by graphing { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Solve the system by graphing: { x − 3 y = −3 x + y = 5 . { x − 3 y = −3 x + y = 5 .

Solve the system by graphing: { − x + y = 1 3 x + 2 y = 12 . { − x + y = 1 3 x + 2 y = 12 .

The steps to use to solve a system of linear equations by graphing are shown here.

Solve a system of linear equations by graphing.

  • Step 1. Graph the first equation.
  • Step 2. Graph the second equation on the same rectangular coordinate system.
  • Step 3. Determine whether the lines intersect, are parallel, or are the same line.
  • If the lines intersect, identify the point of intersection. This is the solution to the system.
  • If the lines are parallel, the system has no solution.
  • If the lines are the same, the system has an infinite number of solutions.
  • Step 5. Check the solution in both equations.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

Example 4.3

Solve the system by graphing: { 3 x + y = − 1 2 x + y = 0 . { 3 x + y = − 1 2 x + y = 0 .

We’ll solve both of these equations for y y so that we can easily graph them using their slopes and y -intercepts.

Solve the system by graphing: { − x + y = 1 2 x + y = 10 . { − x + y = 1 2 x + y = 10 .

Solve the system by graphing: { 2 x + y = 6 x + y = 1 . { 2 x + y = 6 x + y = 1 .

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Example 4.4

Solve the system by graphing: { y = 1 2 x − 3 x − 2 y = 4 . { y = 1 2 x − 3 x − 2 y = 4 .

Solve the system by graphing: { y = − 1 4 x + 2 x + 4 y = − 8 . { y = − 1 4 x + 2 x + 4 y = − 8 .

Solve the system by graphing: { y = 3 x − 1 6 x − 2 y = 6 . { y = 3 x − 1 6 x − 2 y = 6 .

Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.

Example 4.5

Solve the system by graphing: { y = 2 x − 3 − 6 x + 3 y = − 9 . { y = 2 x − 3 − 6 x + 3 y = − 9 .

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.

Solve the system by graphing: { y = − 3 x − 6 6 x + 2 y = − 12 . { y = − 3 x − 6 6 x + 2 y = − 12 .

Try It 4.10

Solve the system by graphing: { y = 1 2 x − 4 2 x − 4 y = 16 . { y = 1 2 x − 4 2 x − 4 y = 16 .

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident . Coincident lines have the same slope and same y- intercept.

Coincident Lines

Coincident lines have the same slope and same y- intercept.

The systems of equations in Example 4.2 and Example 4.3 each had two intersecting lines. Each system had one solution.

In Example 4.5 , the equations gave coincident lines, and so the system had infinitely many solutions.

The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.

A system with parallel lines, like Example 4.4 , has no solution. We call a system of equations like this inconsistent. It has no solution.

Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See below and Table 4.1 .

Example 4.6

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ { y = 3 x − 1 6 x − 2 y = 12 { y = 3 x − 1 6 x − 2 y = 12 ⓑ { 2 x + y = − 3 x − 5 y = 5 { 2 x + y = − 3 x − 5 y = 5

ⓐ We will compare the slopes and intercepts of the two lines.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Try It 4.11

ⓐ { y = −2 x − 4 4 x + 2 y = 9 { y = −2 x − 4 4 x + 2 y = 9 ⓑ { 3 x + 2 y = 2 2 x + y = 1 { 3 x + 2 y = 2 2 x + y = 1

Try It 4.12

ⓐ { y = 1 3 x − 5 x − 3 y = 6 { y = 1 3 x − 5 x − 3 y = 6 ⓑ { x + 4 y = 12 − x + y = 3 { x + 4 y = 12 − x + y = 3

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

Example 4.7

How to solve a system of equations by substitution.

Solve the system by substitution: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.13

Solve the system by substitution: { − 2 x + y = −11 x + 3 y = 9 . { − 2 x + y = −11 x + 3 y = 9 .

Try It 4.14

Solve the system by substitution: { 2 x + y = −1 4 x + 3 y = 3 . { 2 x + y = −1 4 x + 3 y = 3 .

Solve a system of equations by substitution.

  • Step 1. Solve one of the equations for either variable.
  • Step 2. Substitute the expression from Step 1 into the other equation.
  • Step 3. Solve the resulting equation.
  • Step 4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
  • Step 5. Write the solution as an ordered pair.
  • Step 6. Check that the ordered pair is a solution to both original equations.

Be very careful with the signs in the next example.

Example 4.8

Solve the system by substitution: { 4 x + 2 y = 4 6 x − y = 8 . { 4 x + 2 y = 4 6 x − y = 8 .

We need to solve one equation for one variable. We will solve the first equation for y .

Try It 4.15

Solve the system by substitution: { x − 4 y = −4 − 3 x + 4 y = 0 . { x − 4 y = −4 − 3 x + 4 y = 0 .

Try It 4.16

Solve the system by substitution: { 4 x − y = 0 2 x − 3 y = 5 . { 4 x − y = 0 2 x − 3 y = 5 .

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d .

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2 , −2 , we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2 . −2 .

Then rewrite the system of equations.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Example 4.9

How to solve a system of equations by elimination.

Solve the system by elimination: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.17

Solve the system by elimination: { 3 x + y = 5 2 x − 3 y = 7 . { 3 x + y = 5 2 x − 3 y = 7 .

Try It 4.18

Solve the system by elimination: { 4 x + y = − 5 − 2 x − 2 y = − 2 . { 4 x + y = − 5 − 2 x − 2 y = − 2 .

The steps are listed here for easy reference.

Solve a system of equations by elimination.

  • Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
  • Decide which variable you will eliminate.
  • Multiply one or both equations so that the coefficients of that variable are opposites.
  • Step 3. Add the equations resulting from Step 2 to eliminate one variable.
  • Step 4. Solve for the remaining variable.
  • Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
  • Step 6. Write the solution as an ordered pair.
  • Step 7. Check that the ordered pair is a solution to both original equations.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example 4.10

Solve the system by elimination: { 4 x − 3 y = 9 7 x + 2 y = −6 . { 4 x − 3 y = 9 7 x + 2 y = −6 .

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

Try It 4.19

Solve the system by elimination: { 3 x − 4 y = − 9 5 x + 3 y = 14 . { 3 x − 4 y = − 9 5 x + 3 y = 14 .

Try It 4.20

Solve each system by elimination: { 7 x + 8 y = 4 3 x − 5 y = 27 . { 7 x + 8 y = 4 3 x − 5 y = 27 .

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

Example 4.11

Solve the system by elimination: { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 . { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 .

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.

Try It 4.21

Solve each system by elimination: { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 . { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 .

Try It 4.22

Solve each system by elimination: { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 . { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 .

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

Example 4.12

Solve the system by elimination: { 3 x + 4 y = 12 y = 3 − 3 4 x . { 3 x + 4 y = 12 y = 3 − 3 4 x .

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It 4.23

Solve the system by elimination: { 5 x − 3 y = 15 y = − 5 + 5 3 x . { 5 x − 3 y = 15 y = − 5 + 5 3 x .

Try It 4.24

Solve the system by elimination: { x + 2 y = 6 y = − 1 2 x + 3 . { x + 2 y = 6 y = − 1 2 x + 3 .

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Example 4.13

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

Since both equations are in standard form, using elimination will be most convenient.

Since one equation is already solved for y , using substitution will be most convenient.

Try It 4.25

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 4 x − 5 y = −32 3 x + 2 y = −1 { 4 x − 5 y = −32 3 x + 2 y = −1 ⓑ { x = 2 y − 1 3 x − 5 y = −7 { x = 2 y − 1 3 x − 5 y = −7

Try It 4.26

ⓐ { y = 2 x − 1 3 x − 4 y = − 6 { y = 2 x − 1 3 x − 4 y = − 6 ⓑ { 6 x − 2 y = 12 3 x + 7 y = −13 { 6 x − 2 y = 12 3 x + 7 y = −13

Section 4.1 Exercises

Practice makes perfect.

In the following exercises, determine if the following points are solutions to the given system of equations.

{ 2 x − 6 y = 0 3 x − 4 y = 5 { 2 x − 6 y = 0 3 x − 4 y = 5

ⓐ ( 3 , 1 ) ( 3 , 1 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

{ − 3 x + y = 8 − x + 2 y = −9 { − 3 x + y = 8 − x + 2 y = −9

ⓐ ( −5 , −7 ) ( −5 , −7 ) ⓑ ( −5 , 7 ) ( −5 , 7 )

{ x + y = 2 y = 3 4 x { x + y = 2 y = 3 4 x

ⓐ ( 8 7 , 6 7 ) ( 8 7 , 6 7 ) ⓑ ( 1 , 3 4 ) ( 1 , 3 4 )

{ 2 x + 3 y = 6 y = 2 3 x + 2 { 2 x + 3 y = 6 y = 2 3 x + 2 ⓐ ( −6 , 2 ) ( −6 , 2 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

In the following exercises, solve the following systems of equations by graphing.

{ 3 x + y = −3 2 x + 3 y = 5 { 3 x + y = −3 2 x + 3 y = 5

{ − x + y = 2 2 x + y = −4 { − x + y = 2 2 x + y = −4

{ y = x + 2 y = −2 x + 2 { y = x + 2 y = −2 x + 2

{ y = x − 2 y = −3 x + 2 { y = x − 2 y = −3 x + 2

{ y = 3 2 x + 1 y = − 1 2 x + 5 { y = 3 2 x + 1 y = − 1 2 x + 5

{ y = 2 3 x − 2 y = − 1 3 x − 5 { y = 2 3 x − 2 y = − 1 3 x − 5

{ x + y = −4 − x + 2 y = −2 { x + y = −4 − x + 2 y = −2

{ − x + 3 y = 3 x + 3 y = 3 { − x + 3 y = 3 x + 3 y = 3

{ − 2 x + 3 y = 3 x + 3 y = 12 { − 2 x + 3 y = 3 x + 3 y = 12

{ 2 x − y = 4 2 x + 3 y = 12 { 2 x − y = 4 2 x + 3 y = 12

{ x + 3 y = −6 y = − 4 3 x + 4 { x + 3 y = −6 y = − 4 3 x + 4

{ − x + 2 y = −6 y = − 1 2 x − 1 { − x + 2 y = −6 y = − 1 2 x − 1

{ − 2 x + 4 y = 4 y = 1 2 x { − 2 x + 4 y = 4 y = 1 2 x

{ 3 x + 5 y = 10 y = − 3 5 x + 1 { 3 x + 5 y = 10 y = − 3 5 x + 1

{ 4 x − 3 y = 8 8 x − 6 y = 14 { 4 x − 3 y = 8 8 x − 6 y = 14

{ x + 3 y = 4 − 2 x − 6 y = 3 { x + 3 y = 4 − 2 x − 6 y = 3

{ x = −3 y + 4 2 x + 6 y = 8 { x = −3 y + 4 2 x + 6 y = 8

{ 4 x = 3 y + 7 8 x − 6 y = 14 { 4 x = 3 y + 7 8 x − 6 y = 14

{ 2 x + y = 6 − 8 x − 4 y = −24 { 2 x + y = 6 − 8 x − 4 y = −24

{ 5 x + 2 y = 7 − 10 x − 4 y = −14 { 5 x + 2 y = 7 − 10 x − 4 y = −14

{ y = 2 3 x + 1 − 2 x + 3 y = 5 { y = 2 3 x + 1 − 2 x + 3 y = 5

{ y = 3 2 x + 1 2 x − 3 y = 7 { y = 3 2 x + 1 2 x − 3 y = 7

{ 5 x + 3 y = 4 2 x − 3 y = 5 { 5 x + 3 y = 4 2 x − 3 y = 5

{ y = − 1 2 x + 5 x + 2 y = 10 { y = − 1 2 x + 5 x + 2 y = 10

{ 5 x − 2 y = 10 y = 5 2 x − 5 { 5 x − 2 y = 10 y = 5 2 x − 5

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ − 2 x + 2 y = 6 y = −3 x + 1 { − 2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

{ y = −2 x − 1 y = − 1 3 x + 4 { y = −2 x − 1 y = − 1 3 x + 4

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x − 16 y = 8 − x − 8 y = −4 { 2 x − 16 y = 8 − x − 8 y = −4

{ y = 7 8 x + 4 − 7 x + 8 y = 6 { y = 7 8 x + 4 − 7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 − 3 x − y = 0 { 5 x + 2 y = 2 − 3 x − y = 0

{ 6 x − 5 y = −1 2 x + y = 13 { 6 x − 5 y = −1 2 x + y = 13

{ 2 x − 5 y = 7 3 x − y = 17 { 2 x − 5 y = 7 3 x − y = 17

{ 5 x − 3 y = −1 2 x − y = 2 { 5 x − 3 y = −1 2 x − y = 2

{ 3 x − 5 y = −9 5 x + 2 y = 16 { 3 x − 5 y = −9 5 x + 2 y = 16

{ 4 x − 3 y = 3 2 x + 5 y = −31 { 4 x − 3 y = 3 2 x + 5 y = −31

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

{ 11 x + 9 y = −5 7 x + 5 y = −1 { 11 x + 9 y = −5 7 x + 5 y = −1

{ 3 x + 8 y = 67 5 x + 3 y = 60 { 3 x + 8 y = 67 5 x + 3 y = 60

{ 2 x + 9 y = −4 3 x + 13 y = −7 { 2 x + 9 y = −4 3 x + 13 y = −7

{ 1 3 x − y = −3 x + 5 2 y = 2 { 1 3 x − y = −3 x + 5 2 y = 2

{ x + 1 2 y = 3 2 1 5 x − 1 5 y = 3 { x + 1 2 y = 3 2 1 5 x − 1 5 y = 3

{ x + 1 3 y = −1 1 3 x + 1 2 y = 1 { x + 1 3 y = −1 1 3 x + 1 2 y = 1

{ 1 3 x − y = −3 2 3 x + 5 2 y = 3 { 1 3 x − y = −3 2 3 x + 5 2 y = 3

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

{ x − 4 y = −1 − 3 x + 12 y = 3 { x − 4 y = −1 − 3 x + 12 y = 3

{ − 3 x − y = 8 6 x + 2 y = −16 { − 3 x − y = 8 6 x + 2 y = −16

{ 4 x + 3 y = 2 20 x + 15 y = 10 { 4 x + 3 y = 2 20 x + 15 y = 10

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ { 8 x − 15 y = −32 6 x + 3 y = −5 { 8 x − 15 y = −32 6 x + 3 y = −5 ⓑ { x = 4 y − 3 4 x − 2 y = −6 { x = 4 y − 3 4 x − 2 y = −6

ⓐ { y = 7 x − 5 3 x − 2 y = 16 { y = 7 x − 5 3 x − 2 y = 16 ⓑ { 12 x − 5 y = −42 3 x + 7 y = −15 { 12 x − 5 y = −42 3 x + 7 y = −15

ⓐ { y = 4 x + 9 5 x − 2 y = −21 { y = 4 x + 9 5 x − 2 y = −21 ⓑ { 9 x − 4 y = 24 3 x + 5 y = −14 { 9 x − 4 y = 24 3 x + 5 y = −14

ⓐ { 14 x − 15 y = −30 7 x + 2 y = 10 { 14 x − 15 y = −30 7 x + 2 y = 10 ⓑ { x = 9 y − 11 2 x − 7 y = −27 { x = 9 y − 11 2 x − 7 y = −27

Writing Exercises

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

Solve the system of equations by substitution and explain all your steps in words: { 3 x + y = 12 x = y − 8 . { 3 x + y = 12 x = y − 8 .

Solve the system of equations by elimination and explain all your steps in words: { 5 x + 4 y = 10 2 x = 3 y + 27 . { 5 x + 4 y = 10 2 x = 3 y + 27 .

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6

ⓐ by graphing ⓑ by substitution ⓒ Which method do you prefer? Why?

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Course: Algebra 1   >   Unit 4

  • Slope, x-intercept, y-intercept meaning in context
  • Slope and intercept meaning in context
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  • Using slope and intercepts in context
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Solving Systems of Linear Equations Part 1: Using Graphs

Learn how to solve systems of linear equations graphically in this interactive tutorial.

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Attachments

General information, source and access information, aligned standards, suggested tutorials.

Learn to solve word problems represented by systems of linear equations, algebraically and graphically, in this interactive tutorial.

This part 7 in a 7-part series. Click below to explore the other tutorials in the series.

  • Part 1: Solving Systems of Linear Equations: Using Graphs
  • Part 2: Solving Systems of Linear Equations: Substitution
  • Part 3: Solving Systems of Linear Equations: Basic Elimination
  • Part 4: Solving Systems of Linear Equations: Advanced Elimination
  • Part 5: Solving Systems of Linear Equations: Connecting Algebraic Methods to Graphing
  • Part 6: Solving Systems of Linear Equations: Writing Systems from Context

Learn how to create systems of linear equations to represent contextual situations in this interactive tutorial.

This part 6 in a 7-part series. Click below to explore the other tutorials in the series. 

  • Part 7: Solving Systems of Linear Equations: Word Problems (Coming soon)

Learn to solve systems of linear equations by connecting algebraic and graphing methods in this interactive tutorial.

This part 5 in a 7-part series. Click below to explore the other tutorials in the series. 

  • Part 6: Solving Systems of Linear Equations: Writing Systems from Context (Coming soon)

Learn to solve systems of linear equations using advanced elimination in this interactive tutorial.

This part 4 in a 7-part series. Click below to explore the other tutorials in the series.

Part 1: Solving Systems of Linear Equations Part 1: Using Graphs

Part 2: Solving Systems of Linear Equations Part 2: Substitution

  • Part 3: Solving Systems of Linear Equations Part 3: Basic Elimination

Part 5: Solving Systems of Linear Equations Part 5: Connecting Algebraic Methods to Graphing (Coming soon)

Part 6: Solving Systems of Linear Equations Part 6: Writing Systems from Context (Coming soon)

Part 7: Solving Systems of Linear Equations Part 7: Word Problems (Coming soon)

Learn to solve systems of linear equations using basic elimination in this interactive tutorial.

This part 3 in a 7-part series. Click below to explore the other tutorials in the series.

Part 4: Solving Systems of Linear Equations Part 4: Advanced Elimination (Coming soon) Part 5: Solving Systems of Linear Equations Part 5: Connecting Algebraic Methods to Graphing (Coming soon) Part 6: Solving Systems of Linear Equations Part 6: Writing Systems from Context (Coming soon) Part 7: Solving Systems of Linear Equations Part 7: Word Problems (Coming soon)

Learn to solve systems of linear equations using substitution in this interactive tutorial.

This part 2 in a 7-part series. Click below to explore the other tutorials in the series.

Part 3: Solving Systems of Linear Equations Part 3: Basic Elimination (Coming soon)

Part 4: Solving Systems of Linear Equations Part 4: Advanced Elimination (Coming soon)

Follow as we learn why the x -coordinate of the point of intersection of two functions is the solution of the equation f ( x ) = g ( x ) in this interactive tutorial.

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Word Problems and Lines

Lots of word problems can be solved with systems of linear equations. However, before we bother with those, let's look at some word problems that describe single lines, and we're not referring to orderly rows of petrified eighth graders on their way into a school dance.

Going from Words to a Linear Equation

The hardest part of word problems is usually translating from English into math, so we'll practice that part first.

Sample Problem

Write and graph the linear equation described by the following statement:

Jenna works at a retail shop. She makes $10 per hour, plus $3 for each item she sells.

The amount Jenna makes depends on how many items she sells, so our independent variable x should be the number of items she sells, and the dependent variable y should be the amount she's paid. On the graph, the horizontal axis will represent the number of items Jenna sells during one hour, and the vertical axis will represent the amount she gets paid during that hour. By the way, $3 is quite a commission rate, considering that most of the shop's inventory consists of cheap polyester scarves that sell for $10 a pop. Her dad must run the place.

solving linear systems by graphing word problems

Since Jenna is paid $10 if she sells 0 items, the point (0, 10) will be on the graph. If Jenna sells 1 item she's paid $13, and if she sells 2 items she's paid $16, so the points (1, 13) and (2, 16) are also on the graph:

solving linear systems by graphing word problems

Now we can connect the dots with a line:

solving linear systems by graphing word problems

Since it doesn't make sense to have Jenna sell a negative number of items, we only draw the part of the line where x is at least 0. We aren't accounting for the dozens of scarves that Jenna has been sneaking into her bag for herself, in which case she is practically selling a negative number of them, but in the interest of simplicity, let's look the other way and pretend we saw nothing.

From the graph, we can now write an equation for this line in slope-intercept form. The y -intercept is 10, and the slope is 3, so the equation we want is

y = 3 x + 10.

If we want to be fussy, we can also write down the condition

Whenever we're writing an equation for a word problem, we need to specify what the variables are.

Lukas left his house at noon to go for a drive. He drove at a constant speed. In fact, it was so constant we're not entirely convinced he didn't just set a brick on the accelerator, then recline his seat and take a nap. Anyway, Lukas was 200 miles from home at 3 p.m.

Lukas's distance from home depends on how long he's been driving. Let's have x be the number of hours Lukas has been driving, and y be Lukas's distance from home.

The graph set-up will look like this:

solving linear systems by graphing word problems

Since Lukas left from his house, he was 0 miles from his house after 0 hours of driving, so the point (0, 0) is on the graph. After 3 hours of driving, Lukas was 200 miles from his house, so the point (3, 200) is also on the graph. We wonder if all this algebra-talk is distracting Lukas from how homesick he's feeling right about now.

solving linear systems by graphing word problems

Now we connect the dots to make a line:

solving linear systems by graphing word problems

As for the equation, we can see that the slope of the line is

solving linear systems by graphing word problems

Since the y -intercept is 0, the equation of the line is

solving linear systems by graphing word problems

Again, notice that we only graphed half of the line. It doesn't make sense to think about Lukas driving for a negative number of hours, so we leave that part out. He may be speedy, but he's not quite speedy enough to turn back time. No matter how much Cher wishes he could.

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W hy's T his F unny?

Systems of Linear Equations Word Problems - Practice - Expii

Systems of linear equations word problems - practice, explanations (3).

solving linear systems by graphing word problems

(Videos) Set up Word Problems Using a System

by mathman1024

solving linear systems by graphing word problems

This video by mathman1024 works through word problems with systems of equations.

The general guideline to follow when doing these word problems is:

  • Define your variables . Since this is a system, there will be two or more variables.
  • Write your equations. Again, since this is a system, there will be two or more equations. This is also the tricky part which will be focused on in the videos.
  • Solve the system. This can be done using the elimination method , substitution method , or graphing .
  • State your answer.

The first problem that he goes over in the video is, " The sum of two numbers is 79, and their difference is 23. What are the two numbers? "

We want to write algebraic expressions for this system.

Step one is defining the variables. The question asks for two numbers so we can see that these are the variables. Let x=one numbery=the other number The first sentence is, " The sum of two numbers is 79. " Since sum means addition , we write, x+y=79 The next part states, " The difference of those two numbers is 23. " Since difference means subtraction, we write, x−y=23 And now we have our system of equations: {x+y=79x−y=23. This can be solved with substitution but you might notice it would be easier with elimination. x+y=79+x−y=232x+0y=1022x=1022x2=1022x=51 Finally, plug this back in to either equation to find the y value. (51)+y=7951−51+y=79−51y=28 The solution to this word problem's system of equations is (51,28).

The next problem is a little trickier. It isn't as obvious how to set up the equations. It says, " A minor league ballpark attracts 88 fans and draws in $553 in revenue from ticket sales. A child's ticket costs $4 and an adult's ticket is $7. How many of each type of ticket were sold? "

First, we define our variables. The problem asks for how many of each type of ticket. Since there are two types of tickets we can write, Let C=# of childrenA=# of adults The first snippet is, " A minor league ballpark attracts 88 fans ". We know there's a total of 88 fans which are made up of children and adults. So we can write, C+A=88 Next we have, " and draws in $553 in revenue from ticket sales. A child's ticket costs $4 and an adult's ticket is $7. " So we have to factor in the money. We see that we get $4 from each child and we get $7 from each adult. Total, we have $553. So we can write, 4C+7A=553 Now we have the system: {C+A=884C+7A=553. This can be solved by either elimination or substitution. For this one, I personally would choose substitution, but either works.

Let's solve the first equation for C. C+A=88C+A−A=88−AC=88−A Plug this back into the other equation to solve for A. 4(88−A)+7A=553352−4A+7A=553352+3A=553352−352+3A=553−3523A=2013A3=2013A=67 Finally, plug this back into either equation to solve for C. C=88−(67)C=21 Our solution is (21,67).

Remember, the best way to double check your answer to either of these example problems is to plug in your solved coordinates back into the original equations.

Related Lessons

As with most word problems, the most effective approach to take is translation. The goal is to take written information and translate into a new language: math equations.

Let's work through an example to see how this translation works.

solving linear systems by graphing word problems

Image by Clker-Free-Vector-Images via Pixabay ( CC0 )

The key to translating a word problem is to identify the given information. Here's what we know from reading the problem:

  • Cupcakes cost $2
  • Pies cost $7.50
  • William bought 20 items
  • Total cost (pies + cupcakes) = $73

We can get two equations from this information.

First, we know that, all together, William bought 20 items. If we say that the number of cupcakes is C and the number of pies is P, we can write the equation:

solving linear systems by graphing word problems

Systems of Linear Equations: Word Problems

Sometimes, it is helpful to translate a word problem into a system of linear equations and solve the system. After translating words to math, isolate one variable in one equation, then use its corresponding expression to solve for the other variable in the other equation. Finally, solve for the first variable! Here is a graphic with an example.

solving linear systems by graphing word problems

Image source: By Caroline Kulczycky

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Mathematics LibreTexts

1.20: Word Problems for Linear Equations

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Word problems are important applications of linear equations. We start with examples of translating an English sentence or phrase into an algebraic expression.

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

Solution: We call the variable \(x .\) Twice the variable is \(2 x .\) Adding \(2 x\) to 4 gives:

\[4 + 2x\nonumber\]

b) Three times a number is subtracted from 7.

Solution: Three times a number is \(3 x .\) We need to subtract \(3 x\) from 7. This means:\

\[7-3 x\nonumber\]

c) 8 less than a number.

Solution: The number is denoted by \(x .8\) less than \(x\) mean, that we need to subtract 8 from it. We get:

\[x-8\nonumber\]

For example, 8 less than 10 is \(10-8=2\).

d) Subtract \(5 p^{2}-7 p+2\) from \(3 p^{2}+4 p\) and simplify.

Solution: We need to calculate \(3 p^{2}+4 p\) minus \(5 p^{2}-7 p+2:\)

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)\nonumber\]

Simplifying this expression gives:

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)=3 p^{2}+4 p-5 p^{2}+7 p-2 =-2 p^{2}+11 p-2\nonumber\]

e) The amount of money given by \(x\) dimes and \(y\) quarters.

Solution: Each dime is worth 10 cents, so that this gives a total of \(10 x\) cents. Each quarter is worth 25 cents, so that this gives a total of \(25 y\) cents. Adding the two amounts gives a total of

\[10 x+25 y \text{ cents or } .10x + .25y \text{ dollars}\nonumber\]

Now we deal with word problems that directly describe an equation involving one variable, which we can then solve.

Example 18.2

Solve the following word problems:

a) Five times an unknown number is equal to 60. Find the number.

Solution: We translate the problem to algebra:

\[5x = 60\nonumber\]

We solve this for \(x\) :

\[x=\frac{60}{5}=12\nonumber\]

b) If 5 is subtracted from twice an unknown number, the difference is \(13 .\) Find the number.

Solution: Translating the problem into an algebraic equation gives:

\[2x − 5 = 13\nonumber\]

We solve this for \(x\). First, add 5 to both sides.

\[2x = 13 + 5, \text{ so that } 2x = 18\nonumber\]

Dividing by 2 gives \(x=\frac{18}{2}=9\).

c) A number subtracted from 9 is equal to 2 times the number. Find the number.

Solution: We translate the problem to algebra.

\[9 − x = 2x\nonumber\]

We solve this as follows. First, add \(x\) :

\[9 = 2x + x \text{ so that } 9 = 3x\nonumber\]

Then the answer is \(x=\frac{9}{3}=3\)

d) Multiply an unknown number by five is equal to adding twelve to the unknown number. Find the number.

Solution: We have the equation:

\[5x = x + 12.\nonumber\]

Subtracting \(x\) gives

\[4x = 12.\nonumber\]

Dividing both sides by 4 gives the answer: \(x=3\).

e) Adding nine to a number gives the same result as subtracting seven from three times the number. Find the number.

Solution: Adding 9 to a number is written as \(x+9,\) while subtracting 7 from three times the number is written as \(3 x-7\). We therefore get the equation:

\[x + 9 = 3x − 7.\nonumber\]

We solve for \(x\) by adding 7 on both sides of the equation:

\[x + 16 = 3x.\nonumber\]

Then we subtract \(x:\)

\[16 = 2x.\nonumber\]

After dividing by \(2,\) we obtain the answer \(x=8\)

The following word problems consider real world applications. They require to model a given situation in the form of an equation.

Example 18.3

a) Due to inflation, the price of a loaf of bread has increased by \(5 \%\). How much does the loaf of bread cost now, when its price was \(\$ 2.40\) last year?

Solution: We calculate the price increase as \(5 \% \cdot \$ 2.40 .\) We have

\[5 \% \cdot 2.40=0.05 \cdot 2.40=0.1200=0.12\nonumber\]

We must add the price increase to the old price.

\[2.40+0.12=2.52\nonumber\]

The new price is therefore \(\$ 2.52\).

b) To complete a job, three workers get paid at a rate of \(\$ 12\) per hour. If the total pay for the job was \(\$ 180,\) then how many hours did the three workers spend on the job?

Solution: We denote the number of hours by \(x\). Then the total price is calculated as the price per hour \((\$ 12)\) times the number of workers times the number of hours \((3) .\) We obtain the equation

\[12 \cdot 3 \cdot x=180\nonumber\]

Simplifying this yields

\[36 x=180\nonumber\]

Dividing by 36 gives

\[x=\frac{180}{36}=5\nonumber\]

Therefore, the three workers needed 5 hours for the job.

c) A farmer cuts a 300 foot fence into two pieces of different sizes. The longer piece should be four times as long as the shorter piece. How long are the two pieces?

\[x+4 x=300\nonumber\]

Combining the like terms on the left, we get

\[5 x=300\nonumber\]

Dividing by 5, we obtain that

\[x=\frac{300}{5}=60\nonumber\]

Therefore, the shorter piece has a length of 60 feet, while the longer piece has four times this length, that is \(4 \times 60\) feet \(=240\) feet.

d) If 4 blocks weigh 28 ounces, how many blocks weigh 70 ounces?

Solution: We denote the weight of a block by \(x .\) If 4 blocks weigh \(28,\) then a block weighs \(x=\frac{28}{4}=7\)

How many blocks weigh \(70 ?\) Well, we only need to find \(\frac{70}{7}=10 .\) So, the answer is \(10 .\)

Note You can solve this problem by setting up and solving the fractional equation \(\frac{28}{4}=\frac{70}{x}\). Solving such equations is addressed in chapter 24.

e) If a rectangle has a length that is three more than twice the width and the perimeter is 20 in, what are the dimensions of the rectangle?

Solution: We denote the width by \(x\). Then the length is \(2 x+3\). The perimeter is 20 in on one hand and \(2(\)length\()+2(\)width\()\) on the other. So we have

\[20=2 x+2(2 x+3)\nonumber\]

Distributing and collecting like terms give

\[20=6 x+6\nonumber\]

Subtracting 6 from both sides of the equation and then dividing both sides of the resulting equation by 6 gives:

\[20-6=6 x \Longrightarrow 14=6 x \Longrightarrow x=\frac{14}{6} \text { in }=\frac{7}{3} \text { in }=2 \frac{1}{3} \text { in. }\nonumber\]

f) If a circle has circumference 4in, what is its radius?

Solution: We know that \(C=2 \pi r\) where \(C\) is the circumference and \(r\) is the radius. So in this case

\[4=2 \pi r\nonumber\]

Dividing both sides by \(2 \pi\) gives

\[r=\frac{4}{2 \pi}=\frac{2}{\pi} \text { in } \approx 0.63 \mathrm{in}\nonumber\]

g) The perimeter of an equilateral triangle is 60 meters. How long is each side?

Solution: Let \(x\) equal the side of the triangle. Then the perimeter is, on the one hand, \(60,\) and on other hand \(3 x .\) So \(3 x=60\) and dividing both sides of the equation by 3 gives \(x=20\) meters.

h) If a gardener has \(\$ 600\) to spend on a fence which costs \(\$ 10\) per linear foot and the area to be fenced in is rectangular and should be twice as long as it is wide, what are the dimensions of the largest fenced in area?

Solution: The perimeter of a rectangle is \(P=2 L+2 W\). Let \(x\) be the width of the rectangle. Then the length is \(2 x .\) The perimeter is \(P=2(2 x)+2 x=6 x\). The largest perimeter is \(\$ 600 /(\$ 10 / f t)=60\) ft. So \(60=6 x\) and dividing both sides by 6 gives \(x=60 / 6=10\). So the dimensions are 10 feet by 20 feet.

i) A trapezoid has an area of 20.2 square inches with one base measuring 3.2 in and the height of 4 in. Find the length of the other base.

Solution: Let \(b\) be the length of the unknown base. The area of the trapezoid is on the one hand 20.2 square inches. On the other hand it is \(\frac{1}{2}(3.2+b) \cdot 4=\) \(6.4+2 b .\) So

\[20.2=6.4+2 b\nonumber\]

Multiplying both sides by 10 gives

\[202=64+20 b\nonumber\]

Subtracting 64 from both sides gives

\[b=\frac{138}{20}=\frac{69}{10}=6.9 \text { in }\nonumber\]

and dividing by 20 gives

Exit Problem

Write an equation and solve: A car uses 12 gallons of gas to travel 100 miles. How many gallons would be needed to travel 450 miles?

Mathx

  • Linear systems – word problems

When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. Once you do that, these linear systems are solvable just like other linear systems . The same rules apply. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you accustomed to finding elements of linear systems inside of word problems. Example 1 : A farmhouse shelters 16 animals. Some of them are chickens and the others are cows. Altogether these animals have 60 legs. How many chickens and how many cows are in the farmhouse?

First, to make the calculations clearer, we will choose symbols to represent the number of cows and the number of chickens. Let us say that the chickens will be represented with x and the cows with y. Now, this task gave us enough information to make two equations. The first one is that the sum of the number of chickens (x) and the number of cows (y) is 16, since there are only 16 animals in the farmhouse. That equation should look like this: x + y = 16

The second piece of information we have is that the total number of legs in the farmhouse is 60. Since we know that cows have four legs each and chickens have two legs each, we have enough information to make another equation. This one will look like this: 2*x + 4*y = 60

Now we have a system of linear equations with two equations and two variables. The only thing left to do now is to solve the system. We will solve it here for you, but if you need to remind yourself how to do that step by step, read the article called Systems of linear equations . x = 16 – y 32 – 2y + 4y = 60 2y = 28 y = 14 x = 2

We can now see that there are two chickens and 14 cows in the farmhouse. The next example will be a bit harder,

Example 2: Rodney’s Kitchen Supplies makes and sells spoons and forks. It costs the store $2 to buy the supplies needed to make a fork, and $1 for the supplies needed to make a spoon. The store sells the forks for $4 and the spoons for 5$. Last month Rodney’s Kitchen Supplies spent $39 on supplies and sold the all of the forks and spoons that were made last month using those supplies for $93. How many forks and spoons did they make?

As we did in the first example, we will first designate symbols to available variables. So, the number of forks made will be represented with x and the number of spoons with y. Again, we have enough information to make two equations. The total cost of making a particular number of forks (x), which cost $2 to make each, and a particular number of spoons (y), which cost $1 to make each, is $39. So that will be our first equation and it will look like this: 2*x + y = 39 The other piece of information tells us that if we sell that number of forks (x) for $4 each and that number of spoons (y) for $5 each, we will make $93. And that will be our second equation: 4*x + 5*y = 93 This was the hard part. Now all we have to do is to solve this linear system to find how many spoons and how many forks did we make last month. y = 39 – 2x 4x + 5*(39 – 2x) = 93 4x + 195 – 10x = 93 -6x = 93 – 195 -6x = -102 |: (-6) x = 17 y = 5

We can see that last month the store made and sold 17 forks and five spoons.

linear systems problems

Although they can seem complicated, mastery and understanding of linear systems and associated word problems will come with a bit of practice. With experience you will be able to recognize their elements and solve even complicated systems with ease. Feel free to use the math worksheets below to practice solving this type of linear systems.

Linear systems – word problems exams for teachers

Linear systems – word problems worksheets for studets.

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  1. how to solve word problems with linear systems

    solving linear systems by graphing word problems

  2. Solving Linear Systems By Graphing Worksheet

    solving linear systems by graphing word problems

  3. 43 solving linear systems by graphing worksheet

    solving linear systems by graphing word problems

  4. Graphing a System of Linear Equation Word Problem

    solving linear systems by graphing word problems

  5. 30 Solving Linear Systems By Graphing Worksheet

    solving linear systems by graphing word problems

  6. Solving Systems Of Linear Equations By Graphing Worksheet

    solving linear systems by graphing word problems

VIDEO

  1. Solving Linear Systems by Graphing Part 4

  2. Solving Linear Systems by Graphing Part 3

  3. Solving Linear Systems by Graphing

  4. Solving Linear Systems by Graphing Part 2

  5. Solve a System of Equations by graphing Word Problems

  6. Solving Linear Systems by Graphing Grade 9 Academic Lesson 6 7 8 1 14

COMMENTS

  1. Solve a system of equations by graphing: word problems

    Improve your math knowledge with free questions in "Solve a system of equations by graphing: word problems" and thousands of other math skills.

  2. Solving Systems of Equations Word Problems

    Solving Systems of Equations Word Problems Solving Systems of Equations Real World Problems Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method.

  3. 5.1: Solve Systems of Equations by Graphing

    To solve a system of linear equations by graphing. Graph the first equation. Graph the second equation on the same rectangular coordinate system. Determine whether the lines intersect, are parallel, or are the same line. Identify the solution to the system. If the lines intersect, identify the point of intersection.

  4. Systems of Linear Equations Word Problems: Graphing

    Worksheet Systems of Linear Equations Word Problems: Graphing Students practice solving word problems by writing and solving systems of equations in this eighth-grade algebra worksheet! For each problem in this two-page worksheet, students are asked to write and solve a system of equations using the graphing method.

  5. Systems of Linear Equations and Word Problems

    Add Down: $ .01x+.035y=.2$ and solve the system. We can also set up mixture problems with the type of figure below. We add up the terms inside the box, and then multiply the amounts in the boxes by the percentages above the boxes, and then add across. This will give us the two equations.

  6. Systems of Linear Equations Word Problems Solved Graphically

    In this video we learn how to solve a system of linear equations word problem by graphing the equations. We look at examples in both standard and slope-inte...

  7. Linear equations word problems: graphs (practice)

    Algebra 1 Course: Algebra 1 > Unit 4 Lesson 5: Applying intercepts and slope Slope and intercept meaning in context Relating linear contexts to graph features Using slope and intercepts in context Slope and intercept meaning from a table Finding slope and intercepts from tables Linear equations word problems: tables

  8. System-of-Equations Word Problems

    . Practically speaking, this mean that, in each of these points, they have given me values for that make the quadratic equation true. Plugging the three points in the general equation for a quadratic, I get a system of three equations, where the variables stand for the unknown coefficients of that quadratic:

  9. Systems of equations with graphing (practice)

    Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. ... Systems of equations with graphing: y=7/5x-5 & y=3/5x-1. Systems of equations with graphing: exact & approximate solutions ... Setting up a system of linear equations example (weight and price) Creating systems ...

  10. PDF Systems of Equations Word Problems

    Systems of Equations Word Problems Date_____ Period____ 1) Find the value of two numbers if their sum is 12 and their difference is 4. 4 and 8 2) The difference of two numbers is 3. Their sum is 13. Find the numbers. 5 and 8 3) Flying to Kampala with a tailwind a plane averaged 158 km/h. On the return trip the plane only

  11. 5.1 Solve Systems of Equations by Graphing

    Solve a System of Linear Equations by Graphing. In this chapter we will use three methods to solve a system of linear equations. The first method we'll use is graphing. The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines.

  12. Linear systems

    When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. Once you do that, these linear systems are solvable just like other linear systems. The same rules apply.

  13. How to solve systems of linear equations by graphing , examples

    How to solve systems lines (2 variable linear equations) by graphing explained with pictures, examples, and interactive practice problems. 1st you ...

  14. 4.1: Solving Linear Systems by Graphing

    The graphing method for solving linear systems is not ideal when the solution consists of coordinates that are not integers. There will be more accurate algebraic methods in sections to come, but for now, the goal is to understand the geometry involved when solving systems. It is important to remember that the solutions to a system correspond ...

  15. 4.1 Solve Systems of Linear Equations with Two Variables

    Try It 4.6. Solve the system by graphing: { 2 x + y = 6 x + y = 1. In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we'll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

  16. Graphing linear relationships word problems

    Algebra 1 Course: Algebra 1 > Unit 4 Lesson 5: Applying intercepts and slope Using slope and intercepts in context Slope and intercept meaning from a table Finding slope and intercepts from tables Linear equations word problems: tables Linear equations word problems: graphs Linear functions word problem: fuel

  17. Solving Systems of Linear Equations Part 1: Using Graphs

    Part 7: Solving Systems of Linear Equations Part 7: Word Problems (Coming soon) Solving an Equation Using a Graph: Follow as we learn why the x -coordinate of the point of intersection of two functions is the solution of the equation f ( x ) = g ( x ) in this interactive tutorial.

  18. Systems of Linear Equations Word Problems and Lines

    From the graph, we can now write an equation for this line in slope-intercept form. The y -intercept is 10, and the slope is 3, so the equation we want is. y = 3 x + 10. x ≥ 0. Whenever we're writing an equation for a word problem, we need to specify what the variables are.

  19. Solve a system of equations by graphing: word problems

    IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)! Learn more.

  20. Systems of Linear Equations Word Problems

    Systems of Linear Equations: Word Problems. Sometimes, it is helpful to translate a word problem into a system of linear equations and solve the system. After translating words to math, isolate one variable in one equation, then use its corresponding expression to solve for the other variable in the other equation.

  21. 1.20: Word Problems for Linear Equations

    Solution: Translating the problem into an algebraic equation gives: 2x − 5 = 13 2 x − 5 = 13. We solve this for x x. First, add 5 to both sides. 2x = 13 + 5, so that 2x = 18 2 x = 13 + 5, so that 2 x = 18. Dividing by 2 gives x = 182 = 9 x = 18 2 = 9. c) A number subtracted from 9 is equal to 2 times the number.

  22. JMAP A.REI.C.6: Solving Linear Systems, Graphing Linear Systems

    Solve systems of linear equations in two variables both algebraically and graphically. ... Regents-Graphing Linear Systems AI/IA/A/B: 8/4/4/1: TST PDF DOC: AII: Regents-Solving Linear Systems 3 AII/A three variables: 11/1: TST PDF DOC: AI: Practice-Solving Linear Systems 1 substitution: 10: WS PDF:

  23. Linear systems

    When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. Once you do that, these linear systems are solvable just like other linear systems. The same rules apply. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you ...