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• Application of Linear Equations

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There are a number of applications of linear equations in Mathematics and in real life. An algebraic expression consists of variables that are equated to each other using an equal “=” sign, it is called an equation. An equation with a degree of one is termed a linear equation. Mathematical knowledge is usually applied through word problems, and the applications of linear equations are observed on a wide scale to solve such word problems. Here is a detailed discussion of the applications of linear equations and how they will fit in the real world.

Linear Equation

A Linear equation is an algebraic expression that consists of variable and equality sign (=). Linear equations are classified based on the number of variables. 

Linear Equation with One Variable

The equation which has one degree and one variable is called a linear equation with one variable.

For Example: 5x +30 = 0, 3x + 12 = 48

Linear Equation with Two Variable

The equation which has one and two variables is called a linear equation with two variables.

For Example: 2x + 3y =12, 3x + 7y = 42

Representation of Linear Equations

The graphical representation of the linear equation is ax + by + c = 0 . Where a and b are coefficients, x and y are variables, and c is a constant term.

Applications of Linear Equations

The applications of linear equations are vast and are applicable in numerous real-life situations. To handle real-life situations using algebra, we change the given situation into mathematical statements. So that it clearly illustrates the relationship between the unknown variables and the known information. The following are the steps involved to reiterate a situation into a mathematical statement,

Convert the real problem into a mathematical statement and frame it in the form of an algebraic expression that clearly defines the problem situation.

Identify the unknowns in the situation and assign variables of these unknown quantities.

Read the situation clearly a number of times and cite the data, phrases, and keywords. Sequentially organize the obtained information.

Write an equation using the algebraic expression and the provided data in the statement and solve it using systematic equation solving techniques

Reframe the solution to the problem statement and analyze if it exactly suits the problem.

Using these steps, the applications of word problems can be solved easily.

Applications of Linear Equations in Real life

The following are some of the examples in which applications of linear equations are used in real life.

It can be used to solve age related problems.

It is used to calculate speed, distance and time of a moving object.

Geometry related problems can be solved.

It is used to calculate money and percentage related problems.

Work, time and wages problems can be solved.

Problems based on force and pressure can be solved.

FAQs on Application of Linear Equations

1. How can we Calculate the Slope of a Line using Linear Equations?

The equation for the slope of a line is y = mx + c. here, x and y are the variables, c is the constant term, and “m” is the slope of the line. Using trial and error methods, we can find solutions for x and y points and then we will have only one unknown in our equation which is “m”, and we can find that by substituting the x and y values.

2. What is the Elimination Method of Solving Linear equations and When can it be used?

The elimination method of solving linear equations can be used when solving a system of two linear equations which have more than one variable.

This process involves the multiplication of one equation by a constant and then adding or subtracting the equations in order to get an equation in one variable. This can then be solved by the substitution of values to get the final solution. For problems with this kind of situation elimination of solving linear equations is used.

3. What is a linear equation? How is it classified?

A Linear equation is an algebraic expression that consists of variable and equality sign (=). The graphical representation of the linear equation is ax + by + c = 0. Where a and b are coefficients, x and y are variables, and c is a constant term.

Linear equations are classified based on the number of variables. 

4. What are the real life applications of linear equations?

It can be used to calculate the age of a person in the past, present, and future.

It is used to calculate the speed, distance, and time of a moving object.

It can be used to solve geometrical problems such as lines, parabolas, etc.

It is used to calculate money and percentage-related problems.

Work, time, and wages problems can be solved.

Problems based on the force and pressure of a pipe can be solved.

5. What are the steps to be considered while converting real life problems into linear equations?

6. How to calculate the slope of a line using linear equations?

The equation for the slope of a line is represented as y = mx + c. here, x and y are the variables, c is the constant term, and “m” is the slope of the line. Using the trial and error methods, we can find solutions for x and y points, and then we will have only one unknown in our equation which is “m”, and we can find that by substituting the x and y values.

Applications of Linear Equations

Applications of linear equations are used by people on a daily basis even without using a line graph because the situations faced by them might have an unknown quantity that can be represented as a linear equation such as calculating mileage rates, income over time, etc. The important aspect here is the four main arithmetic operations - addition, subtraction, multiplication, and division. Let us learn more about the applications of linear equations in this article and solve some real-life examples.

What is the Application of Linear Equations?

The main objective for the applications of linear equations or linear systems is to solve various problems using two variables where one is known and the other is unknown, also dependent on the first. Some of these applications of linear equations are:

• Geometry problems by using two variables.
• Money problems by using two variables
• A mixture of problems by using two variables.
• Distance-Rate-Time problems by using two variables.
• Application of linear equation in business and economics

Definition of Linear Equations

An equation with degree 1 is considered a linear equation. In other words, a linear equation is an equation that is written for two different variables. This equation will be a linear combination of these two variables, and a constant can be present. The standard form is ax + by = c, where a and b are coefficients , x and y are variables, and c is the constant. A linear equation can also be plotted on a graph where it is a straight line. The image below shows a few examples of linear equations.

Real-Life Applications of Linear Equations

Applications of linear equations are required almost frequently in our life even if that means doing calculations mentally at times. And this is where the first application of linear equation comes. Some of the real-life applications of linear equations could be calculating the cost of hiring a taxi on vacation, it could be a useful tool to compare the better rates of payment for work or budgeting or making any sort of predictions. Let us look at an example to understand this better. Let us find electricity consumed on days 1,2,3 and so on. Every day, the average amount of electricity consumed would be 23 units. So, electricity consumed at the end of the nth day is X = 23 × n + 0.

Let us look at another example. Suppose we rent a car with a charge of $200 plus $25 for every hour. Here you don’t know how many hours you will travel so by using "t" to represent the number of hours to your destination and "x" to represent the cost of that taxi ride, this can be framed in an equation as x = 25 × t + 200.

Application of Linear Equations in Word Problems

Application of linear equations is used in solving word or story problems which can be difficult in the beginning but once the process is understood, solving word problems will become simple. The process is as follows:

Read the Problem Again

You must read the problem carefully to understand what is given, what kind of application of linear system it is, and what is to be found. If required read it as many times as it takes. It is imperative for you to completely understand and identify all the given information and what you are being asked to find. Be mindful as at times one word can completely change how the application of a linear system is worked.

Assign Variables

Represent one of the unknown values/quantity with a variable and try to relate all the other unknown values (if any) to this variable, using diagrams or tables, etc. Pen down what each variable represents.

Write Equations

Form equations that relate unknown quantities to known quantities. We can use some of the known formulas and often the figure/equations sketched in the previous step can be used to determine the relevant equation leading to answer what we need to find.

Solve the System of Equations

Solve the equations formed in the previous step and answer all the questions asked as we may be asked several values but the equation will only give one of them.

Check Answers

You can check the answer by plugging it into the equation but also ensure that it makes sense. Sometimes mistakes can be identified by acknowledging that the answer does not make sense.

There is some standard formula that is used in some of the word problems, they are:

• Distance=Rate × Time
• Work Done = Work Rate × Time Spent Working
• Amount of secondary liquid in the water = Percentage of Solution × Volume of Solution

Related Topics 

Listed below are a few topics related to applications of linear equations, take a look. 

• Standard Form of Linear Equations
• Solving Linear Equations 
• Graphing Linear Equations Calculator

Examples on Applications of Linear Equations

Example 1: Five years ago, the age of Niraj was seven times that of his son Girish. Five years from, his age will be three times that of his son. What is the ratio of their ages today?

Solution: Let the ages of Niraj and Girish be x and y respectively. Then, their ages five years ago would have been:

(x - 5), (y - 5)

And their ages five years from today will be:

(x + 5) , (y + 5)

From the information given in the problem, we have:

\[\begin{array}{l}\left\{ \begin{array}{l}x - 5 = 7\left( {y - 5} \right)\\x + 5 = 3\left( {y + 5} \right)\end{array} \right.\\ \Rightarrow \qquad x - 7y + 30 = 0\\ \qquad\quad x - 3y - 10 = 0\end{array}\]

Subtracting the two equations, we have:

\[\begin{array}{l} - 4y + 40 = 0\\ \Rightarrow y = 10\end{array}\]

Plugging this value of y back into the first equation, we have:

\[\begin{array}{l}x - 5 = 7\left( {y - 5} \right)\\ \Rightarrow \qquad x - 5 = 7\left( {10 - 5} \right) = 35\\ \Rightarrow \qquad x = 40\end{array}\]

Thus, their present ages are 40 years and 10 years.

Example 2: Solve the following pair of equations:

\[\begin{align}&\frac{2}{x} + \frac{3}{y} = 18\\&\frac{5}{x} - \frac{2}{y} = 7\end{align}\]

Solution: Clearly, this is not a pair of linear equations, but it can easily be reduced to one, using the following substitutions:

\[\frac{1}{x} \to p,\;\frac{1}{y} \to q\]

Thus, our pair of equations becomes:

\[\begin{array}{l}\left\{ \begin{array}{l}2p + 3q = 18\\5p - 2q = 7\end{array} \right.\\ \Rightarrow \;\;\;2p + 3q - 18 = 0\\\;\;\;\;\;\;5p - 2q - 7 = 0\end{array}\]

Let’s use cross-multiplication to solve this pair:

\[\begin{align}&\frac{p}{{\left( {3 \times - 7} \right) - \left( { - 2 \times - 18} \right)}} = \frac{{ - q}}{{\left( {2 \times - 7} \right) - \left( {5 \times - 18} \right)}}\\ &\qquad\qquad\qquad\qquad\qquad\;\;= \frac{1}{{\left( {2 \times - 2} \right) - \left( {5 \times 3} \right)}}\\ &\Rightarrow \;\;\;\frac{p}{{ - 21 - 36}} = \frac{{ - q}}{{ - 14 + 90}} = \frac{1}{{ - 4 - 15}}\\ &\Rightarrow \;\;\;\frac{p}{{ - 57}} = \frac{{ - q}}{{76}} = \frac{1}{{ - 19}}\\ &\Rightarrow \;\;\;p = 3,\;q = 4\end{align}\]

Note that these values correspond to p and q , whereas the original unknowns were x and y , so we still have one more step left:

\[\begin{align}{l}x = \frac{1}{p} = \frac{1}{3}\\y = \frac{1}{q} = \frac{1}{4}\end{align}\]

This completes our solution.

Example 3: Solve the following pair of equations:

\[\begin{align}{l}\frac{5}{{x - 1}} + \frac{1}{{y - 2}} = 2\\\frac{6}{{x - 1}} - \frac{3}{{y - 2}} = 1\end{align}\]

Solution: Once again, these equations are not linear, but we can easily reduce them to linear form:

\[\frac{1}{{x - 1}} \to p,\;\frac{1}{{y - 2}} \to q\]

Thus, our equations become:

\[\begin{array}{l}\left\{ \begin{array}{l}5p + q = 2\\6p - 3q = 1\end{array} \right.\\ \Rightarrow \;\;\;5p + q - 2 = 0\\\;\;\;\;\;\;6p - 3q - 1 = 0\end{array}\]

By cross-multiplication, the solution is:

\[\begin{align}&\frac{p}{{\left( {1 \times - 1} \right) - \left( { - 3 \times - 2} \right)}} = \frac{{ - q}}{{\left( {5 \times - 1} \right) - \left( {6 \times - 2} \right)}}\\ &\qquad\qquad\qquad\qquad\qquad= \frac{1}{{\left( {5 \times - 3} \right) - \left( {6 \times 1} \right)}}\\ &\Rightarrow \;\;\;\frac{p}{{ - 7}} = \frac{{ - q}}{7} = \frac{1}{{ - 21}}\\ &\Rightarrow \;\;\;p = \frac{1}{3},\;q = \frac{1}{3}\end{align}\]

Now, we evaluate the original unknowns:

\[\begin{align}&\frac{1}{{x - 1}} = p = \frac{1}{3}\\ &\Rightarrow \;\;\;x - 1 = 3\;\; \Rightarrow \;\;x = 4\\&\frac{1}{{y - 2}} = q = \frac{1}{3}\\ &\Rightarrow \;\;\;y - 2 = 3\;\; \Rightarrow \;\;y = 5\end{align}\]

Thus, the final solution is:

\[x = 4,\;y = 5\]

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Practice Questions on Applications of Linear Equations

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FAQs on Applications of Linear Equations

What are the applications of linear equations in real life.

In real-life situations where there is an unknown quantity or identity, the use of linear equations comes into play, for example, figuring out income over time, calculating mileage rates, or predicting profit. Most of the time mental calculations are used in some real-life situations without drawing a line graph .

What is a Linear Equation? Explain with an Example.

An equation of the form ax + by = c is called a linear equation. Here, x and y are variables, and a, b and c are constants. Examples of the linear equation are:

• 7y - 5x = 1

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

• Substitution method
• Elimination method
• Cross multiplication method
• Graphical method

What are the Applications of Linear Equations in Two Variables?

The application of linear equations in two variables is solving equations involving two different variables. 

How Do you Graph a Linear Equation?

The basic methods of the graphing linear equation:

• The first method is by plotting all points on the graph and then drawing a line through the points.
• The second is by using the y-intercept of the equation and the slope of the equation.

What is the General Form of a linear equation in One Variable?

The general form of the linear equations in one variable is Ax + B = 0 where x is the variable and A is the coefficient of x and b is the constant term.

• Math Article
• Applications Of Linear Equations

Linear Equations : Applications

There are various applications of linear equations in Mathematics as well as in real life. An algebraic equation is an equality that includes variables and equal sign (=). A linear equation is an equation of degree one.

The knowledge of mathematics is frequently applied through word problems, and the applications of linear equations are observed on a wide scale to solve such word problems. Here, we are going to discuss the linear equation applications and how to use them in the real world with the help of an example.

What is Linear Equation?

A linear equation is an algebraic expression with a variable and equality sign (=), whose highest degree is equal to 1. For example, 2x – 1 = 5 is a linear equation.

• A linear equation with one variable and degree one is called a linear equation in one variable. (Eg, 3x + 5 = 0)
• A linear equation with degrees one and two variables is called a linear equation in two variables. (Eg, 3x + 5y = 0)

The graphical representation of linear equation is ax + by + c = 0, where,

• a and b are coefficients
• x and y are variables
• c is a constant term

What are the Applications of Linear Equations?

In real life, the applications of linear equations are vast. To tackle real-life problems using algebra, we convert the given situation into mathematical statements in such a way that it clearly illustrates the relationship between the unknowns (variables) and the information provided. The following steps are involved while restating a situation into a mathematical statement:

• Translate the problem statement into a mathematical statement and set it up in the form of algebraic expression in a manner it illustrates the problem aptly.
• Identify the unknowns in the problem and assign variables ( quantity whose value can change depending upon the mathematical context ) to these unknown quantities.
• Read the problem thoroughly multiple times and cite the data, phrases and keywords. Organize the information obtained sequentially.
• Frame an equation with the help of the algebraic expression and the data provided in the problem statement and solve it using systematic techniques of equation solving.
• Retrace your solution to the problem statement and analyze if it suits the criterion of the problem.

There you go!! Using these steps and applications of linear equations word problems can be solved easily.

Applications of Linear equations in Real life

• Finding unknown age
• Finding unknown angles in geometry
• For calculation of speed, distance or time
• Problems based on force and pressure

Let us look into an example to analyze the applications of linear equations in depth.

Applications of Linear Equations Solved Example

Rishi is twice as old as Vani. 10 years ago his age was thrice of Vani. Find their present ages.

In this word problem, the ages of Rishi and Vani are unknown quantities. Therefore as discussed above, let us first choose variables for the unknowns.

Let us assume that Vani’s present age is ‘x’ years. Since Rishi’s present age is 2 times that of Vani, therefore his present age can be assumed to be ‘2x’.

10 years ago, Vani’s age would have been ‘x – 10 ’, and Rishi’s age would have been ‘2x – 10’. According to the problem statement, 10 years ago, Rishi’s age was thrice of Vani, i.e. 2x – 10 = 3(x – 10).

We have our linear equation in the variable ‘x’ which clearly defines the problem statement. Now we can solve this linear equation easily and get the result.

This implies that the current age of Vani is 20 years, and Rishi’s age is ‘2x,’ i.e. 40 years. Let us retrace our solution. If the present age of Vani is 20 years then 10 years ago her age would have been 10 years, and Rishi’s age would have been 30 years which satisfies our problem statement. Thus, applications of linear equations enable us to tackle such real-world problems.

Video Lesson on Applications of Linear Equation

Related Articles

• Solving Linear Equations
• Algebra – Linear Equations Applications
• Difference Between Linear and Nonlinear Equations
• Graphing of Linear Equations
• Linear Equation in One Variable
• Linear Equations In Two Variables

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Module 1: Algebraic Expressions, Linear Equations and Mathematical Models

1.5 – applications of linear equations, learning objectives, translate words into algebraic expressions and equations.

• Solve an application using a formula

(1.5.2) – Solve distance, rate, and time problems

(1.5.3) – solve area and perimeter problems, (1.5.4) – rearrange formulas to isolate specific variables, (1.5.5) – solve temperature conversion problems.

Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus an additional charge if you exceed your data plan; a car rental company charges a daily fee plus an amount per mile driven; Chipotle offers a base price for your burrito plus additional charges for extra toppings, like guacamole and sour cream. 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.

(1.5.1) – Set Up a Linear Equation to Solve an 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 use an example of a car rental company. The company charges $0.10/mi in addition to a flat rate. 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 [latex]0.10x[/latex]. 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 [latex]C[/latex].

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

How To: 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.

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

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

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 [latex]A=0.05x+34[/latex]. This includes the variable cost of [latex]0.05x[/latex] plus the monthly service charge of $34. Company B ’s package charges a higher monthly fee of $40, but a lower variable cost of [latex]0.04x[/latex]. Company B ’s model can be written as [latex]B=0.04x+\$40[/latex].

Check the x- value in each equation.

Solve an Application Using a Formula

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, [latex]A=LW[/latex]; the perimeter of a rectangle, [latex]P=2L+2W[/latex]; and the volume of a rectangular solid, [latex]V=LWH[/latex]. 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.

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 [latex]d=rt[/latex], 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 [latex]\frac{1}{2}[/latex] h at rate [latex]r[/latex]. His drive home takes 40 min, or [latex]\frac{2}{3}[/latex] h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance [latex]d[/latex]. A table, such as the one below, 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 [latex]r[/latex].

[latex]\begin{array}{l}r\left(\frac{1}{2}\right)\hfill&=\left(r - 10\right)\left(\frac{2}{3}\right)\hfill \\ 6\times r\left(\frac{1}{2}\right)\hfill& =6\times \left(r - 10\right)\left(\frac{2}{3}\right)\hfill \\ 3r\hfill& =4\left(r - 10\right)\hfill \\ 3r\hfill& =4r - 40\hfill \\ -r\hfill& =-40\hfill \\ r\hfill& =40\hfill \end{array}[/latex]

In the next example, we will find the lengths of a rectangular field given it’s perimeter and a relationship between it’s side lengths.

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

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

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

The dimensions are [latex]L=15[/latex] ft and [latex]W=12[/latex] ft.

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

The standard formula for area is [latex]A=LW[/latex]; 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 [latex]L=W+6[/latex]. 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 [latex]L=15[/latex] in. and [latex]W=9[/latex] in.

The following video shows another example of finding the area of a rectangle given it’s perimeter and the relationship between it’s side lengths.

Sometimes, it is easier to isolate the variable you you are solving for when you are using a formula. This is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly. In the next examples, we will use algebraic properties to isolate a variable in a formula.

Isolate the term containing the variable, w, from the formula for the perimeter of a rectangle :  

[latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex].

First, isolate the term with  w by subtracting 2 l from both sides of the equation.

[latex] \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,p\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\p-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}[/latex]

Next, clear the coefficient of w by dividing both sides of the equation by 2.

[latex]\displaystyle \begin{array}{l}\underline{p-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\frac{p-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\frac{p-2l}{2}\end{array}[/latex]

You can rewrite the equation so the isolated variable is on the left side.

[latex]w=\frac{p-2l}{2}[/latex]

Use the multiplication and division properties of equality to isolate the variable b  given [latex]A=\frac{1}{2}bh[/latex]

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}[/latex]

Write the equation with the desired variable on the left-hand side as a matter of convention:

Use the multiplication and division properties of equality to isolate the variable h  given [latex]A=\frac{1}{2}bh[/latex]

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}[/latex]

Let’s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.

[latex]C=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Given a temperature of [latex]12^{\circ}{C}[/latex], find the equivalent in [latex]{}^{\circ}{F}[/latex].

Substitute the given temperature in[latex]{}^{\circ}{C}[/latex] into the conversion formula:

[latex]12=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Isolate the variable F to obtain the equivalent temperature.

[latex]\begin{array}{r}12=\left(F-32\right)\cdot \frac{5}{9}\\\\\left(\frac{9}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\frac{108}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.

To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \displaystyle \frac{9}{5}[/latex].

[latex]\begin{array}{l}\\\,\,\,\,\left(\frac{9}{5}\right)C=\left(F-32\right)\left(\frac{5}{9}\right)\left(\frac{9}{5}\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\frac{9}{5}C=F-32\end{array}[/latex]

Add 32 to both sides.

[latex]\begin{array}{l}\frac{9}{5}\,C+32=F-32+32\\\\\frac{9}{5}\,C+32=F\end{array}[/latex]

[latex]F=\frac{9}{5}C+32[/latex]

• College Algebra. Authored by : Abramson, Jay, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution . License Terms : Download for free at : http://cnx.org/contents/[email protected]:1/Preface
• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• Write and Solve Linear Equation - Number Problem with Given Sum. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/CGC_SwHfwMk . License : CC BY: Attribution
• Write Linear Equations to Model and Compare Cell Phone Plans with Data Usage. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Provided by : https://youtu.be/Q5hlC_VPKGM. License : CC BY: Attribution
• Ex: Find the Dimensions and Area of a Field Given the Perimeter. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/VyK-HQr02iQ . License : CC BY: Attribution
• Ex: Find the Area of a Rectangle Given the Perimeter. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/zUlU64Umnq4 . License : CC BY: Attribution
• Find the Volume of a Right Circular Cylinder Formed from a Given Rectangle. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/KS1pGO_g3vM . License : CC BY: Attribution

MA001: College Algebra

Applications of Linear Equations

This section provides you with applications of the linear equation and its representation on the Cartesian plane. Examples are given in the context of real-world models and scenarios.

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

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.

EXAMPLE 1

Modeling a linear equation to solve an unknown number problem, try it #1, example 2, 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?

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

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?

Mathematics PreCalculus Mathematics at Nebraska

Josh Brummer, Taran Funk, Lawrence Seminario-Romero, Ariel Setniker, Karina Uhing, Nathan Wakefield, Catherine Zimmitti,

Section Applications of Linear Equations

Algebra simplifies the process of solving real-world problems. This is done by using letters to represent unknowns, restating problems in the form of equations, and by offering systematic techniques for solving those equations. In this section, we will solve several types of real world problems using algebra.

In this section, you will...

translate verbal descriptions into equations

write and solve linear equations to represent real world applications

Supplemental Videos

The main topics of this section are also presented in the following videos:

• Translate Verbal Descriptions into Equations, A Few Examples

To solve problems using algebra, first translate the wording of the problem into mathematical statements that describe the relationships between the given information and the unknowns. Usually, this translation to mathematical statements is the difficult step in the process. The key to the translation is to carefully read the problem and identify certain key words and phrases.

When translating sentences into mathematical statements, be sure to read the sentence several times and parse out the key words and phrases. It is important to first identify what needs to be represented by a variable, and then specify what that variable will be. For instance, if we are solving a word problem that asks us to find the distance that a car travelled, we likely will want to define a variable to represent the distance the car travelled. It is important to be specific and include units if applicable: it is better to say "\(d=\)the distance travelled by the car in miles" than simply saying "\(d=\)distance".

In the previous paragraph, we chose the variable \(d\) to represent distance. You can choose any letter for your variables, but it can be helpful to chose a letter that aligns with what the variable represents (such as \(d\) for distance, \(t\) for time, etc.). However, we often just use \(x\text{.}\) As long as you clearly define your variables, it does not matter what variables you use.

Before we work with word problems, we will first practice translating simple sentences into equations.

Write an equation for the sequence of operations to describe each situation below.

\(x\) is \(9\) more than \(y\text{.}\)

Half of \(x\) is \(y\text{.}\)

\(x\) is \(30\%\) of \(y\text{.}\)

In all of these examples, it can be helpful to think of the word "is" as the equals sign.

\(x\) is \(9\) more than \(y\) means that \(x=y+9\text{.}\)

Half of \(x\) is \(y\) means that \(\frac{1}{2}x=y\text{.}\)

\(x\) is \(30\%\) of \(y\) means that \(x=0.30y\text{.}\)

We will now expand these ideas to solve a word problem using a linear equation.

When 6 is subtracted from twice the sum of a number and 8, the result is 5. Find the number.

Let \(n\) represent the unknown number. In the sentence above, "the sum of a number and 8" is represented by \(n+8\text{,}\) so "twice the sum of a number and 8" is \(2(n+8)\text{.}\) Next, "6 subtracted from twice the sum of a number and 8" would be 6 less than this, so it is \(2(n+8)-6\text{.}\) Finally, we are told that the result is 5; in other words, this expressions is equal to 5:

Now that we have translated the problem into a linear equation, we can solve for \(n\text{:}\)

We should always check our answer to be sure that we didn't make a mistake in setting up or solving the problem. The sum of \(-\frac{5}{2}\) and \(8\) is \(-\frac{5}{2}+8=\frac{11}{2}\text{.}\) Twice this is \(\frac{11}{2}\cdot2=11\text{,}\) and \(6\) less than this is \(5\text{.}\) Since we got the desired result, we know that our answer is correct. The unknown numbers is \(-\frac{5}{2}\text{.}\)

General guidelines for setting up and solving word problems follow:

Guidelines for Word Problems

• Read the problem several times, identify the key words and phrases, and organize the given information.
• Identify the variables by assigning a letter or expression to the unknown quantities.
• Translate and set up an algebraic equation that models the problem.
• Solve the resulting algebraic equation.
• Finally, answer the question in sentence form and make sure it makes sense (check it).

All of the equations in the following examples can be set up using only one variable. We can avoid two variables by looking for a relationship between the unknowns.

A rectangle has a perimeter measuring \(92\) meters. The length is \(2\) meters less than \(3\) times the width. Find the dimensions of the rectangle.

The sentence "The length is \(2\) meters less than \(3\) times the width" gives us the relationship between the two important quantities, the length and the width. Let \(w\) represent the width of the rectangle. Since the length is \(2\) meters less than \(3\) times the width, the length must be \(3w-2\text{.}\)

The sentence "A rectangle has a perimeter measuring \(92\) meters" suggests an algebraic set up. The perimeter of a rectangle is always the sum of twice the length and twice the width; in this example, this means that the perimeter is

Since we know that the perimeter is \(92\) meters, we have the following equation:

Once you have set up an algebraic equation with one variable, solve for the width, \(w\text{.}\)

Now that we know the width, we can use the fact that the length is \(3w-2\) to find the length.

This means that the dimensions of the rectangle are \(12\) meters by \(34\) meters. To check, we make sure the perimeter is actually \(92\) meters:

For the rest of this section, we will explore several different categories of application problems that you might see, such as applications of interest, currency, distances, and more.

Subsection Interest Applications

In problems involving simple interest, it is often helpful to use the simple interest formula \(I=prt\) where \(I\) is the simple interest earned on a principle investment, \(p\text{,}\) with an interest rate of \(r\) over \(t\) years.

Given a \(4.375\)% annual interest rate, how long will it take \($2,500\) to yield \($437.50\) in simple interest?

Let \(t\) represent the time needed to earn $\(437.50\) at \(4.375\)%. Organize the information needed to use the formula for simple interest, \(I=prt\text{.}\)

Next, substitute all of the known quantities into the formula and then solve for the only unknown, \(t\text{.}\)

It takes \(4\) years for \($2,500\) invested at \(4.375\)% to earn \($437.50\) in simple interest.

Stephanie invested her total savings of \($12,500\) in two accounts earning simple interest. Her mutual fund account earned \(7\)% last year and her CD earned \(4.5\)%. If her total interest for the year was \($670\text{,}\) how much was in each account?

The sum of the money in each account is \($12,500\text{.}\) When a total is involved, a common technique used to avoid two variables is to represent the second unknown as the difference of the total and the first unknown.

Let \(m\) represent the amount invested in the mutual fund.

Since the amount in the two accounts total \($12,500\text{,}\) the amount in the CD must be \(12,500-m\text{.}\)

The total interest is the sum of the interest earned from each account.

This equation models the problem with one variable. Solve for \(m\text{.}\)

Use \(12,500-m\) to find the amount in the CD.

Stephanie invested \($4,300\) at \(7\)% in a mutual fund and \($8,200\) at \(4.5\)% in a CD.

Subsection Currency Applications

A new kind of problem we come across in applications is something we call currency problems. These problems involve finding the number of coins or bills given other information about value. It is important to be careful in defining our variables specifically - either for the number of currency or the value of the currency.

Ashley has $17.25 in her piggy bank, which has only quarters and fifty-cent pieces in it. Ashley has nine more fifty-cent pieces than she has quarters. How many coins of each denomination does she have?

Let \(q\) represent the number of quarters. Since Ashley has nine more fifty-cent pieces than quarters, the number of fifty cent pieces she has is \(q+9\text{.}\)

Since a quarter is worth \($0.25\text{,}\) \(0.25q\) represents the value of Ashley's quarters. Similarly, since a fifty-cent piece is worth \($0.50\text{,}\) \(0.50(q+9)\) gives the value of Ashley's fifty-cent pieces.

Since Ashley has a total of \($17.25\text{,}\) it must be the case that the sum of these values is \(17.25\text{:}\)

We can now solve for \(q\text{:}\)

Ashley has \(17\) quarters, and therefore \(26\) fifty-cent pieces.

Subsection Mixing Applications

Mixture problems often include a percentage and some total amount. It is important to make a distinction between these two types of quantities. For example, if a problem states that a 10-ounce container is filled with a \(3\)% saline (salt) solution, then this means that the container is filled with a mixture of salt and water as follows:

In other words, we multiply the percentage times the total to get the amount of each part of the mixture.

How many milliliters of a \(15\)% alcohol solution must be mixed with \(110\) milliliters of a \(35\)% solution to obtain a \(25\)% solution?

It is often helpful to use a table to organize your work for application problems. We will use the following table to keep track of all the different solutions in this problem:

Let \(m\) be the milliliters of the \(15\)% solution. To get the amount of alcohol in this part of the mixture, we multiply the percentage by the amount: \(0.15m\text{.}\)

Likewise, to get the amount of the second part of the mixture, we multiply the percentage by the amount: \(0.35\cdot 110=38.5\text{.}\)

Let's now think about the mixture of these two solutions. We know that we're aiming for a \(25\)% solution. Since there are \(m\) milliliters from the first solution and \(110\) milliliters from the second solution, the mixture will have \(m+110\) milliliters in it. Again, we find the amount of alcohol in the mixture by multiplying the percentage by the amount: \(0.25(m+110)\text{.}\) We can now fill in our table.

From our table, we see that the total amount of alcohol in the mixture is \(0.25(m+110)\text{.}\) On the other hand, we know that there are \(0.15m\) milliliters from the first solution and \(38.5\) milliliters from the second solution, for a total of \(0.15m+38.5\) milliliters of alcohol in the solution. We have found two different expressions for the total amount of alcohol in the mixture, so they must be equal:

We can now solve for \(m\text{.}\)

We have found that we must mix \(110\) milliliters of a \(15\)% alcohol solution with \(110\) milliliters of a \(35\)% solution to obtain a \(25\)% solution.

Subsection Distance Applications

The distance traveled is equal to the average rate times the time traveled at that rate, \(d=r\cdot t\text{.}\) These problems usually have a lot of data, so it helps to carefully rewrite the pertinent information as you define variables.

Two planes leave a city traveling in opposite directions. One travels at a rate of \(530\) mph and the other at a rate of \(600\) mph. How long will it take until they are \(3672.5\) miles apart?

Let \(d_1\) represent the distance travelled by the first plane in miles, and let \(d_1\) represent the distance travelled by the second plane in miles. The key to this problem is that the sum of their distances is the distance between them:

Let \(t\) represent the time it takes for them to be \(3672.5\) miles apart (in hours); this is what we're actually looking for. The first plane travels at a rate of \(530\) mph, meaning that its distance is modeled by the equation

The second plane travels at a rate of \(600\) mph, meaning that its distance is modeled by the equation

Since the sum of their distances is the distance between them, after \(t\) hours the two planes will be

miles apart. Now, we want to know when the distance is \(3672.5\text{.}\) Thus we solve:

This tells us that it will take \(3\) hours and \(15\) minutes for the planes to be \(3672.5\) miles apart.

Subsection Exercises

1 translating verbal to equations, 2 translating verbal to equations, 3 automobile journey, 4 mario and luigi, 5 average speed, 6 mixing solutions, 9 class schedule, 10 simple interest, 11 sales tax, 12 loose change.