# 2.2: Linear Equations in One Variable with Applications

- Page ID
- 152029

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- Solve linear equations in one variable using properties of equations.
- Construct a linear equation to solve applications.
- 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

$ax+b=0$

where $a$ and $b$ are real numbers and $a\ne 0$, 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. We will follow the steps below in examples 2.2.1 and 2.2.2.

### How To

#### Solve a Linear Equation

- Simplify each side of the equation as much as possible.
- Collect all variable terms on one side of the equation.
- Collect constant terms on the other side of the equation.
- Make the coefficient of the variable term equal to 1.
- Check the solution.

### Example 2.2.1

#### Solving a Linear Equation Using a General Strategy

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

**Answer**-
**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. 7 (n - 3) - 8 = -15

7n - 21 - 8 = -15

7n - 29 = -15

**Step 2:**Collect all variable terms on one side of the equation.Nothing to do; all $n$-terms are on the left side. 7n - 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.

7n - 29

**+29**= -15**+29**7n = 14

**Step 4:**Make the coefficient of the variable term equal to 1.Divide each side by 7.

Simplify.

$\frac{7n}{7}=\frac{14}{7}$

n = 2

**Step 5:**Check the solution.Let $n=2$

Subtract.

Check:

$\begin{array}{ccc}\hfill 7(n-3)-8& \hfill =\hfill & -15\hfill \\ \hfill 7(2-3)-8& \hfill \stackrel{?}{=}\hfill & -15\hfill \\ \hfill 7\left(-1\right)-8& \hfill \stackrel{?}{=}\hfill & -15\hfill \\ \hfill -7-8& \hfill \stackrel{?}{=}\hfill & -15\hfill \\ \hfill -15& \hfill =\hfill & -15\hfill \end{array}$

### Your Turn 2.2.1

In Example 2.2.1, 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 $a+c=b+c$ |
$\begin{array}{ccc}\hfill 2& \hfill =\hfill & 2\hfill \\ \hfill 2+3& \hfill =\hfill & 2+3\hfill \\ \hfill 5& \hfill =\hfill & 5\hfill \end{array}$ |

Subtraction |
If $a=b$ Then $a-c=b-c$ |
$\begin{array}{ccc}\hfill 5& \hfill =\hfill & 5\hfill \\ \hfill 5-2& \hfill =\hfill & 5-2\hfill \\ \hfill 3& \hfill =\hfill & 3\hfill \end{array}$ |

Multiplication |
If $a=b$ Then $a\u2022c=b\u2022c$ |
$\begin{array}{ccc}\hfill 3& \hfill =\hfill & 3\hfill \\ \hfill 3\u20224& \hfill =\hfill & 3\u20224\hfill \\ \hfill 12& \hfill =\hfill & 12\hfill \end{array}$ |

Division |
If $a=b$ Then $a\xf7c=b\xf7c$ for $c\ne 0$ |
$\begin{array}{ccc}\hfill 8& \hfill =\hfill & 8\hfill \\ \hfill 8\xf72& \hfill =\hfill & 8\xf72\hfill \\ \hfill 4& \hfill =\hfill & 4\hfill \end{array}$ |

### Example 2.2.2

#### Solving a Linear Equation Using Properties of Equations

Solve 9 (y - 2) - y = 16 + 7y.

**Answer**-
**Step 1:**Simplify each side.9 (y - 2) - y = 16 + 7y

9y - 18 - y = 16 + 7y

8y - 18 = 16 +7y

**Step 2:**Collect all variables on one side by subtracting 7y from both sides.8y - 18

**- 7y**= 16 + 7y**- 7y**y - 18 = 16 **Step 3:**Collect constant terms on one side by adding 18 to both sides.y - 18

**+ 18**= 16**+ 18**y = 34

**Step 4:**Make the coefficient of the variable 1. Already done!**Step 5:**Check.$$\begin{array}{ccc}\hfill 9\left(34\right)-18-\left(34\right)& \hfill \stackrel{?}{=}\hfill & 16+7\left(34\right)\hfill \\ \hfill 306-18-34& \hfill \stackrel{?}{=}\hfill & 16+238\hfill \\ \hfill 288-34& \hfill \stackrel{?}{=}\hfill & 254\hfill \\ \hfill 254& \hfill =\hfill & 254\u2713\hfill \end{array}$$

### Your Turn 2.2.2

### Who Knew?

#### Who Invented the Symbol for Equals ?

Before the creation of a symbol for equality, it was usually expressed with a word that meant equals, such as *aequales* (Latin), *esgale* (French), or *gleich* (German). Welsh mathematician and physician Robert Recorde is given credit for inventing the modern sign. It first appears in writing in *The Whetstone of Witte*, a book Recorde wrote about algebra, which was published in 1557. In this book, Recorde states, "I will set as I do often in work use, a pair of parallels, or Gemowe (twin) lines of one length, thus: ===, because no two things can be more equal." Although his version of the sign was a bit longer than the one we use today, his idea stuck and "=" is used throughout the world to indicate equality in mathematics.

Working with percents in Sections 1.5 and 1.6, 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 2.2.4).

### Example 2.2.3

#### Constructing a Linear Equation to Solve an Application

The Beaudrie family has two cats, Basil and Max. Together, they weigh 23 pounds. Basil weighs 16 pounds. How much does Max weigh?

**Answer**-
**Steps 1 and 2:**Let $b$ = Basil’s weight and $m$ = Max’s weight.

$$b+m=23$$

We also know that Basil weighs 16 pounds so replace b with 16 in the equation.

$16+m=23$

**Step 3:**Since both sides are simplified and the variable is on one side of the equation, we start solving the equation by collecting the constants on one side.$$\begin{array}{ccc}\hfill 16+m-\mathbf{16}& \hfill =\hfill & 23-\mathbf{16}\hfill \\ \hfill m& \hfill =\hfill & 7\hfill \end{array}$$

**Step 4:**is already done, so we go to Step 5.**Step 5:**$$\begin{array}{ccc}\hfill 16+7& \hfill \stackrel{?}{=}\hfill & 23\hfill \\ \hfill 23& \hfill =\hfill & 23\u2713\hfill \end{array}$$

Basil weighs 16 pounds and Max weighs 7 pounds.

### Your Turn 2.2.3

### Example 2.2.4

#### 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?

**Answer**-
If we let $x$ = number of classes, the expression $5x+$$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 $5x+10=75$ for $x$.

**Steps 1 and 2:**$$5x+10=75$$

**Step 3:**$$\begin{array}{ccc}\hfill 5x+\mathbf{10}-10& \hfill =\hfill & 75-\mathbf{10}\hfill \\ \hfill 5x& \hfill =\hfill & 65\hfill \end{array}$$

**Step 4:**$$\begin{array}{ccc}\hfill \frac{5x}{\mathbf{5}}& \hfill =\hfill & \frac{65}{\mathbf{5}}\hfill \\ \hfill x& \hfill =\hfill & 13\hfill \end{array}$$

**Step 5:**$$\begin{array}{ccc}\hfill 5\left(13\right)+10& \hfill \stackrel{?}{=}\hfill & 75\hfill \\ \hfill 65+10& \hfill \stackrel{?}{=}\hfill & 75\hfill \\ \hfill 75& \hfill =\hfill & \hfill 75\u2713\hfill \end{array}$$

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

### Your Turn 2.2.4

### Example 2.2.5

#### Constructing an Application from a Linear Equation

Write an application that can be solved using the equation $$

**Answer**-
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 + 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 $$50 x + 35 = 185 for $x$.**Steps 1 and 2:**$$50x+35=185$$

**Step 3:**$$\begin{array}{ccc}\hfill 50x+35\mathbf{-}\mathbf{35}& \hfill =\hfill & 185\mathbf{-}\mathbf{35}\hfill \\ \hfill 50x& \hfill =\hfill & 150\hfill \end{array}$$

**Step 4:**$$\begin{array}{ccc}\hfill \frac{50x}{\mathbf{50}}& \hfill =\hfill & \frac{150}{\mathbf{50}}\hfill \\ \hfill x& \hfill =\hfill & 3\hfill \end{array}$$

**Step 5:**$$\begin{array}{ccc}\hfill 50\left(3\right)+35& \hfill \stackrel{?}{=}\hfill & 185\hfill \\ \hfill 150+35& \hfill \stackrel{?}{=}\hfill & 185\hfill \\ \hfill 185& \hfill =\hfill & 185\u2713\hfill \end{array}$$

The equation is $$

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

### Your Turn 2.2.5

### 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.

### Example 2.2.6

#### Solving for a Given Variable with Distance, Rate, and Time

Solve the formula $d=rt$ for $t$. This is the distance formula where d stands for distance, r stands for rate, and t stands for time.

**Answer**-
Divide both sides by $r$: $d/\mathit{r}=rt/\mathit{r}$

Simplify.$$d/r=t$$

### Your Turn 2.2.6

### Video

### Example 2.2.7

#### Solving for a Given Variable in the Area Formula for a Triangle

Solve the formula $A=$$\frac{1}{2}$$bh$ for $h$. This is the area formula of a triangle where $A$ = area, $b$ = base, and $h$ = height.

**Answer**-
**Step 1:**Multiply both sides by 2.$$\begin{array}{l}2A=2(\mathrm{\xbd}\phantom{\rule{0.28em}{0ex}}bh)\hfill \\ 2A=bh\hfill \end{array}$$

**Step 2:**Divide both sides by $b$.$$\begin{array}{ccc}\hfill \frac{2A}{b}& \hfill =\hfill & \frac{bh}{b}\hfill \\ \hfill \frac{2A}{b}& \hfill =\hfill & h\hfill \\ \hfill \end{array}$$