5.2 Linear Equations in One Variable with Applications

A man is lifting weights at a gym. — **Figure 5.4** Most gyms have a monthly membership fee.Most gyms have a monthly membership fee. (credit: modification of work "Morning PT after the Holidays 2021" by Fort Drum & 10th Mountain Division (LI)/Flickr, Public Domain Mark 1.0)

Learning Objectives

After completing this section, you should be able to:

Solve linear equations in one variable using properties of equations.
Construct a linear equation to solve applications.
Determine equations with no solution or infinitely many solutions.
Solve a formula for a given variable.

In this section, we will study linear equations in one variable. There are several real-world scenarios that can be represented by linear equations: taxi rentals with a flat fee and a rate per mile; cell phone bills that charge a monthly fee plus a separate rate per text; gym memberships with a monthly fee plus a rate per class taken; etc. For example, if you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month?

Linear Equations and Applications

Solving any equation is like discovering the answer to a puzzle. The purpose of solving an equation is to find the value or values of the variable that makes the equation a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle! There are many types of equations that we will learn to solve. In this section, we will focus on a linear equation, which is an equation in one variable that can be written as

$a x + b = 0$

where $a$ and $b$ are real numbers and $a not equal to 0$ , such that $a$ is the coefficient of $x$ and $b$ is the constant.

To solve a linear equation, it is a good idea to have an overall strategy that can be used to solve any linear equation. In the Example 1, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

Example 1 Solving a Linear Equation Using a General Strategy

Solve $7 (n - 3) - 8 = - 15$

Step 1: Simplify each side of the equation as much as possible.	Use the Distributive Property. Notice that each side of the equation is now simplified as much as possible.	$\begin{matrix} 7 (n - 3) - 8 & = & - 15 \\ 7 n - 21 - 8 & = & - 15 \\ 7 n - 29 & = & - 15 \end{matrix}$
Step 2: Collect all variable terms on one side of the equation.	Nothing to do; all $n$ -terms are on the left side.	$7 n - 29 = - 15$
Step 3: Collect constant terms on the other side of the equation.	To get constants only on the right, add 29 to each side. Simplify.	$\begin{matrix} 7 n - 29 + 29 & = & - 15 + 29 \\ 7 n & = & 14 \end{matrix}$
Step 4: Make the coefficient of the variable term equal to 1.	Divide each side by 7. Simplify.	$\frac{7 n}{7} = \frac{14}{7}$ $n equals 2$
Step 5: Check the solution.	Let $n equals 2$ Subtract.	Check: $\begin{matrix} 7 (n - 3) - 8 & = & - 15 \\ 7 (2 - 3) - 8 & \overset{?}{=} & - 15 \\ 7 (- 1) - 8 & \overset{?}{=} & - 15 \\ - 7 - 8 & \overset{?}{=} & - 15 \\ - 15 & = & - 15 ✓ \end{matrix}$

In Example 1, we used both the addition and division property of equations. All the properties of equations are summarized in table below. Basically, what you do to one side of the equation, you must do to the other side of the equation to preserve equality.

Operation	Property	Example
Addition	If $a = b$ Then $a + c = b + c$	$\begin{matrix} 2 & = & 2 \\ 2 + 3 & = & 2 + 3 \\ 5 & = & 5 \end{matrix}$
Subtraction	If $a = b$ Then $a - c = b - c$	$\begin{matrix} 5 & = & 5 \\ 5 - 2 & = & 5 - 2 \\ 3 & = & 3 \end{matrix}$
Multiplication	If $a = b$ Then $a \cdot c = b \cdot c$	$\begin{matrix} 3 & = & 3 \\ 3 \cdot 4 & = & 3 \cdot 4 \\ 12 & = & 12 \end{matrix}$
Division	If $a = b$ Then $a \div c = b \div c$ for $c not equal to 0$	$\begin{matrix} 8 & = & 8 \\ 8 \div 2 & = & 8 \div 2 \\ 4 & = & 4 \end{matrix}$

In Algebraic Expressions, you translated an English sentence into an equation. In this section, we take that one step further and translate an English paragraph into an equation, and then we solve the equation. We can go back to the opening question in this section: If you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month? We can create an equation for this scenario and then solve the equation (see Example 4).

Example 4 Constructing a Linear Equation to Solve Another Application

If you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month?

If we let $x$ = number of classes, the expression $5 x plus 10$ would represent what you pay per month if each class is $5 and there’s a $10 monthly fee per class. $10 is your constant. If you want to know how many classes you can take if you have a $75 monthly gym budget, set the equation equal to 75. Then solve the equation $5 x + 10 = 75$ for $x$ .

Steps 1 and 2:

$5 x + 10 = 75$

Step 3:

$\begin{matrix} 5 x + 10 - 10 & = & 75 - 10 \\ 5 x & = & 65 \end{matrix}$

Step 4:

$\begin{matrix} \frac{5 x}{5} & = & \frac{65}{5} \\ x & = & 13 \end{matrix}$

Step 5:

$\begin{matrix} 5 (13) + 10 & \overset{?}{=} & 75 \\ 65 + 10 & \overset{?}{=} & 75 \\ 75 & = & 75 ✓ \end{matrix}$

The solution is 13 classes. You can take 13 classes on a $75 monthly gym budget.

Example 5 Constructing an Application from a Linear Equation

Write an application that can be solved using the equation $50 x + 35 = 185$ . Then solve your application.

Answers will vary. Let’s say you want to rent a snowblower for a huge winter storm coming up. If $x$ = the number of days you rent a snowblower, then the expression $50 x plus 35$ represents what you pay if, for each day, it costs $50 to rent the snowblower and there is a $35 flat rental fee. $35 is the constant. To find out how many days you can rent a snowblower for $185, set the expression equal to 185. Then solve the equation $50 x + 35 = 185$ for $x$ .

Steps 1 and 2:

$50 x + 35 = 185$

Step 3:

$\begin{matrix} 50 x + 35 - 35 & = & 185 - 35 \\ 50 x & = & 150 \end{matrix}$

Step 4:

$\begin{matrix} \frac{50 x}{50} & = & \frac{150}{50} \\ x & = & 3 \end{matrix}$

Step 5:

$\begin{matrix} 50 (3) + 35 & \overset{?}{=} & 185 \\ 150 + 35 & \overset{?}{=} & 185 \\ 185 & = & 185 ✓ \end{matrix}$

The equation is $50 x + 35 = 185$ and the solution is 3 days. You can rent a snowblower for 3 days on a $185 budget.

Linear Equations with No Solutions or Infinitely Many Solutions

Every linear equation we have solved thus far has given us one numerical solution. Now we'll look at linear equations for which there are no solutions or infinitely many solutions.

Solving a Formula for a Given Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be able to manipulate formulas and solve for specific variables.

To solve a formula for a specific variable means to isolate that variable on one side of the equal sign with a coefficient of 1. All other variables and constants are on the other side of the equal sign. To see how to solve a formula for a specific variable, we will start with the distance, rate, and time formula.

Work It Out

Using Algebra to Understand Card Tricks

You will need to perform this card trick with another person. Before you begin, the two people must first decide which of the two will be the Dealer and which will be the Partner, as each will do something different. Once you have decided upon that, follow the steps here:

Step 1: Dealer and Partner: Take a regular deck of 52 cards, and remove the face cards and the 10s.

Step 2: Dealer and Partner: Shuffle the remaining cards

Step 3: Dealer and Partner: Select one card each, but keep them face down and don’t look at them yet.

Step 4: Dealer: Look at your card (just the Dealer!). Multiply its value by 2 (Aces = 1).

Step 5: Dealer: Add 2 to this result.

Step 6: Dealer: Multiply your answer by 5.

Step 7: Partner: Look at your card.

Step 8: Partner: Calculate: 10 - your card, and tell this information to the dealer.

Step 9: Dealer: Subtract the value the Partner tells you from your total to get a final answer.

Step 10: Dealer: verbally state the final answer.

Step 11: Dealer and Partner: Turn over your cards. Now, answer the following questions

Did the trick work? How do you know?
Why did this occur? In other words, how does this trick work?

Key Terms

linear equation

Key Concepts

Solving linear equations means discovering what the value of the variable in a linear equation represents in the given conditions.
When solving a linear equation, most often you will have one solution; however, a linear equation may have no solutions or infinitely many solutions.

Videos

Solving for a Variable in an Equation ↗ new tab

Adapted from Contemporary Mathematics by OpenStax (openstax.org), licensed under CC BY-NC-SA 4.0. Changes were made. License: CC-BY-NC-SA-4.0.