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 realworld 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
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. In the Example 5.12, 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 5.12
Solving a Linear Equation Using a General Strategy
Solve $7(n3)8=15$
Solution
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.  $\begin{array}{ccc}\hfill 7(n3)8& \hfill =\hfill & 15\hfill \\ \hfill 7n218& \hfill =\hfill & 15\hfill \\ \hfill 7n29& \hfill =\hfill & 15\hfill \end{array}$ 
Step 2: Collect all variable terms on one side of the equation.  Nothing to do; all $n$terms are on the left side.  $7n29=15$ 
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. 
$\begin{array}{ccc}\hfill 7n\mathbf{29}+29& \hfill =\hfill & 15+\mathbf{29}\hfill \\ \hfill 7n& \hfill =\hfill & 14\hfill \end{array}$ 
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: $\begin{array}{ccc}\hfill 7(n3)8& \hfill =\hfill & 15\hfill \\ \hfill 7(23)8& \hfill \stackrel{?}{=}\hfill & 15\hfill \\ \hfill 7\left(1\right)8& \hfill \stackrel{?}{=}\hfill & 15\hfill \\ \hfill 78& \hfill \stackrel{?}{=}\hfill & 15\hfill \\ \hfill 15& \hfill =\hfill & 15\u2713\hfill \end{array}$ 
Your Turn 5.12
In Example 5.12, 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 $ac=bc$ 
$\begin{array}{ccc}\hfill 5& \hfill =\hfill & 5\hfill \\ \hfill 52& \hfill =\hfill & 52\hfill \\ \hfill 3& \hfill =\hfill & 3\hfill \end{array}$ 
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 $c\ne 0$ 
$\begin{array}{ccc}\hfill 8& \hfill =\hfill & 8\hfill \\ \hfill 8\xf72& \hfill =\hfill & 8\xf72\hfill \\ \hfill 4& \hfill =\hfill & 4\hfill \end{array}$ 
Checkpoint
Be careful to multiply and divide every term on each side of the equation. For example, $2+x=\frac{x}{3}$ is solved by multiplying BOTH sides of the equation by 3 to get $3\left(2+x\right)=3\left(\frac{x}{3}\right)$ which gives $6+3x=x$. Using parentheses will help you remember to use the distributive property! A division example, such as $3\left(x+2\right)=6x+9$, can be solved by dividing BOTH sides of the equation by 3 to get $\frac{3(x+2)}{3}=\frac{6x+9}{3},$ which then will lead to $x+2=2x+3$.
Example 5.13
Solving a Linear Equation Using Properties of Equations
Solve $9\left(y2\right)y=16+7y$.
Solution
Step 1: Simplify each side.
Step 2: Collect all variables on one side.
Step 3: Collect constant terms on one side.
Step 4: Make the coefficient of the variable 1. Already done!
Step 5: Check.
Your Turn 5.13
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.
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 5.15).
Example 5.14
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?
Solution
Let $b$ = Basil’s weight and $m$ = Max’s weight.
We also know that Basil weighs 16 pounds so:
Steps 1 and 2: $16+m=23$
Since both sides are simplified, the variable is on one side of the equation, we start in Step 3 and collect the constants on one side:
Step 3:
Step 4: is already done so we go to Step 5:
Step 5:
Basil weighs 16 pounds and Max weighs 7 pounds.
Your Turn 5.14
Example 5.15
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?
Solution
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:
Step 3:
Step 4:
Step 5:
The solution is 13 classes. You can take 13 classes on a $75 monthly gym budget.
Your Turn 5.15
Example 5.16
Constructing an Application from a Linear Equation
Write an application that can be solved using the equation $50x+35=185$. Then solve your application.
Solution
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 $50x+35$ 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 $50x+35=185$ for $x$.
Steps 1 and 2:
Step 3:
Step 4:
Step 5:
The equation is $50x+35=185$ and the solution is 3 days. You can rent a snowblower for 3 days on a $185 budget.
Your Turn 5.16
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.
Example 5.17
Solving a Linear Equation with No Solution
Solve $3\left(x+4\right)=4x+8x$.
Solution
Step 1: Simplify each side. $3\left(x+4\right)=4x+8x$
Step 2: Collect all variables to one side. $3x+12\mathbf{}\mathbf{3}\mathit{x}=3x+8\mathbf{}\mathbf{3}\mathit{x}$
The variable $x$ disappeared! When this happens, you need to examine what remains. In this particular case, we have $12=8$, which is not a true statement. When you have a false statement, then you know the equation has no solution; there does not exist a value for $x$ that can be put into the equation that will make it true.
Your Turn 5.17
Example 5.18
Solving a Linear Equation with Infinitely Many Solutions
Solve $2\left(x+5\right)=4\left(x+3\right)2x2$.
Solution
Step 1:
Step 2:
As with the previous example, the variable disappeared. In this case, however, we have a true statement ($10=10$). When this occurs we say there are infinitely many solutions; any value for $x$ will make this statement true.
Your Turn 5.18
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 5.19
Solving for a Given Variable with Distance, Rate, and Time
Solve the formula $d=rt$ for $t$. This is the distance formula where $d$ = distance, $r$ = rate, and $t$ = time.
Solution
Divide both sides by $r$: $d/\mathit{r}=rt/\mathit{r}$
$$d/r=t$$
Your Turn 5.19
Example 5.20
Solving for a Given Variable in the Area Formula for a Triangle
Solve the formula $A=\mathrm{\xbd}$ $\mathit{bh}$ for $h$. This is the area formula of a triangle where $A$ = area, $b$ = base, and $h$ = height.
Solution
Step 1: Multiply both sides by 2.
Step 2: Divide both sides by $b$.
Your Turn 5.20
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?
Check Your Understanding
{8\left( {x2} \right)}&{ = }&{6\left( {x + 10} \right)}\\{8x16}&{ = }&{6x + 60}\\{8x16{\mathbf\,{\,6x}}}&{ = }&{6x + 60{\mathbf\,{\,6x}}}\\{2x16 + {\mathbf{16}}}&{ = }&{60 + {\mathbf{16}}}\\{2x}&{ = }&{76}\\{x}&{ = }&{38}\end{array}
\begin{array}{rcl}{7 + 4\left( {2 + 5x} \right)}&{ = }&{3\left( {6x + 7} \right)\left( {13x + 36} \right)}\\{7 + 8 + 20x}&{ = }&{18x + 2113x36}\\{15 + 20x}&{ = }&{5x15}\\{15 + 20x\,{\mathbf{\,5\mathit{x}}}}&{ = }&{5x15\,{\mathbf{\,5\mathit{x}}}}\\{15 + 15x\,{\mathbf{\,15}}}&{ = }&{ 15\,{\mathbf{\,15}}}\\{15x}&{ = }&{ \,30}\\{x}&{ = }&{ \,2}\end{array}
\begin{array}{rcl}{8x + 7\left( {2x9} \right)}&{ = }&{22\left( {4x4} \right)}\\{8x + 72x9}&{ = }&{224x4}\\{6x2}&{ = }&{184x}\\{6x2 +{\mathbf{ 4\mathit{x}}}}&{ = }&{184x +{\mathbf{4\mathit{x}}}}\\{10x2 +{\mathbf{ 2}}}&{ = }&{18 +{\mathbf{ 2}}}\\{10x}&{ = }&{20}\\{x}&{ = }&{2}\end{array}

F = \frac{5}{9}C32

F = \frac{9}{5}C32

F = \frac{5}{9}C + 32

F = \frac{9}{5}C + 32

C = K + 273

K = C + 273

K = C273

C = K273

K = \frac{5}{9}\left( {F32} \right) + 273

K = \frac{5}{9}F + 241

K = \frac{9}{5}\left( {F32} \right) + 273

K = \frac{9}{5}F + 241

R = \frac{5}{9}C492

R = \frac{9}{5}C + 492

R = C + 492

R = \frac{5}{9}\left( {C492} \right)