Learning Objectives

In this section you will:

Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of$400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 1. 2.2.1 – Solving Linear Equations in One Variable A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form $\,ax+b=0\,$ and are solved using basic algebraic operations. We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation. $3x=2x+x$ The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for $\,x\,$ will make the equation true. A conditional equation is true for only some values of the variable. For example, if we are to solve the equation $\,5x+2=3x-6,$ we have the following: $\begin{array}{ccc}\hfill 5x+2& =& 3x-6\hfill \\ \hfill 2x& =& -8\hfill \\ \hfill x& =& -4\hfill \end{array}$ The solution set consists of one number: $\,\left\{-4\right\}.\,$ It is the only solution and, therefore, we have solved a conditional equation. An inconsistent equation results in a false statement. For example, if we are to solve $\,5x-15=5\left(x-4\right),$ we have the following: $\begin{array}{cccc}\hfill 5x-15& =& 5x-20\hfill & \\ \hfill 5x-15-5x& =& 5x-20-5x\hfill & \phantom{\rule{2em}{0ex}}\text{Subtract }5x\text{ from both sides}.\hfill \\ \hfill -15& \ne & -20\hfill & \phantom{\rule{2em}{0ex}}\text{False statement}\hfill \end{array}$ Indeed, $\,-15\ne \,-20.\,$ There is no solution because this is an inconsistent equation. Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows. Linear Equation in One Variable A linear equation in one variable can be written in the form $ax+b=0$ where a and b are real numbers, $\,a\ne 0.$ How To Given a linear equation in one variable, use algebra to solve it. The following steps are used to manipulate an equation and isolate the unknown variable. There is no set order, as the steps used depend on what is given: 1. We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero. 2. Apply the distributive property as needed: $\,a\left(b+c\right)=ab+ac.$ 3. Isolate the variable on one side of the equation. 4. When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient. Example 1 – Solving an Equation in One Variable Solve the following equation: $\,2x+7=19.$ This equation can be written in the form $\,ax+b=0\,$ by subtracting $\,19\,$ from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations. $\begin{array}{cccc}\hfill 2x+7& =& 19\hfill & \\ \hfill 2x& =& 12\hfill & \phantom{\rule{2em}{0ex}}\text{Subtract 7 from both sides}\text{.}\hfill \\ \hfill x& =& 6\hfill & \phantom{\rule{2em}{0ex}}\text{Multiply both sides by }\frac{1}{2}\text{ or divide by 2}\text{.}\hfill \end{array}$ The solution is 6 Try It Solve the linear equation in one variable: $\,2x+1=-9.$ Show answer $x=-5$ Example 2 – Solving an Equation Algebraically When the Variable Appears on Both Sides Solve the following equation: $\,4\left(x-3\right)+12=15-5\left(x+6\right).$ Apply standard algebraic properties. $\begin{array}{cccc}\hfill 4\left(x-3\right)+12& =& 15-5\left(x+6\right)\hfill & \\ 4x-12+12\hfill & =& 15-5x-30\hfill & \phantom{\rule{2em}{0ex}}\text{Apply the distributive property}\text{.}\hfill \\ \hfill 4x& =& -15-5x\hfill & \phantom{\rule{2em}{0ex}}\text{Combine like terms}.\hfill \\ \hfill 9x& =& -15\hfill & \phantom{\rule{2em}{0ex}}\text{Place }x-\text{terms on one side and simplify}.\hfill \\ \hfill x& =& -\frac{15}{9}\hfill & \phantom{\rule{2em}{0ex}}\text{Multiply both sides by }\frac{1}{9}\text{, the reciprocal of 9}.\hfill \\ \hfill x& =& -\frac{5}{3}\hfill & \end{array}$ Analysis This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, $\,x=-\frac{5}{3}.$ Try It Solve the equation in one variable: $\,-2\left(3x-1\right)+x=14-x.$ Show answer $x=-3$ 2.2.2 – Setting up a Linear Equation to Solve a Real-World Application Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer. Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. 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. 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 us use the car rental example above. 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 $\,0.10x.\,$ 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 $\,C.$

$C=0.10x+50$

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

Verbal Translation to Math Operations
One number exceeds another by a $x,\text{​}\,x+a$
Twice a number $2x$
One number is a more than another number $x,\text{​}\,x+a$
One number is a less than twice another number $x,\,2x-a$
The product of a number and a, decreased by b $ax-b$
The quotient of a number and the number plus a is three times the number $\frac{x}{x+a}=3x$
The product of three times a number and the number decreased by b is c $3x\left(x-b\right)=c$

How To

Given a real-world problem, model a linear equation to fit it.

1. Identify known quantities.
2. Assign a variable to represent the unknown quantity.
3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
4. Write an equation interpreting the words as mathematical operations.
5. Solve the equation. Be sure the solution can be explained in words, including the units of measure.

Example 3 – Modeling a Linear Equation to Solve an Unknown Number Problem

Find a linear equation to solve for the following unknown quantities: One number exceeds another number by $\,17\,$ and their sum is $\,31.\,$ Find the two numbers.

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

$\begin{array}{ccc}\hfill x+\left(x+17\right)& =& 31\hfill \\ \hfill 2x+17& =& 31\phantom{\rule{2em}{0ex}}\text{Simplify and solve}\text{.}\hfill \\ \hfill 2x& =& 14\hfill \\ \hfill x& =& 7\hfill \\ \phantom{\rule{1em}{0ex}}\hfill x+17& =& 7+17\hfill \\ & =& 24\hfill \end{array}$

The two numbers are $\,7\,$ and $\,24.$

Try It

Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is $\,36,$ find the numbers.

11 and 25

Example 4 – 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.

1. Write a linear equation that models the packages offered by both companies.
2. If the average number of minutes used each month is 1,160, which company offers the better plan?
3. If the average number of minutes used each month is 420, which company offers the better plan?
4. How many minutes of talk-time would yield equal monthly statements from both companies?
1. The model for Company A can be written as $\,A=0.05x+34.\,$ This includes the variable cost of $\,0.05x\,$ plus the monthly service charge of $34. Company B’s package charges a higher monthly fee of$40, but a lower variable cost of $\,0.04x.\,$ Company B’s model can be written as $\,B=0.04x+\text{\}40.$
2. If the average number of minutes used each month is 1,160, we have the following:

$\begin{array}{ccc}\hfill \text{Company }A& =& 0.05\left(1.160\right)+34\hfill \\ & =& 58+34\hfill \\ & =& 92\hfill \\ \phantom{\rule{1em}{0ex}}\hfill \text{Company }B& =& 0.04\left(1,1600\right)+40\hfill \\ & =& 46.4+40\hfill \\ \hfill & =& 86.4\hfill \end{array}$

So, Company B offers the lower monthly cost of $86.40 as compared with the$92 monthly cost offered by Company A when the average number of minutes used each month is 1,160.

3. If the average number of minutes used each month is 420, we have the following:$\begin{array}{ccc}\hfill \text{Company }A& =& 0.05\left(420\right)+34\hfill \\ & =& 21+34\hfill \\ & =& 55\hfill \\ \phantom{\rule{1em}{0ex}}\hfill \text{Company }B& =& 0.04\left(420\right)+40\hfill \\ & =& 16.8+40\hfill \\ & =& 56.8\hfill \end{array}$

If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B’s monthly cost of$56.80.

4. To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of $\,\left(x,y\right)\,$ coordinates: At what point are both the x-value and the y-value equal? We can find this point by setting the equations equal to each other and solving for x.$\begin{array}{ccc}\hfill 0.05x+34& =& 0.04x+40\hfill \\ \hfill 0.01x& =& 6\hfill \\ \hfill x& =& 600\hfill \end{array}$

Check the x-value in each equation.

$\begin{array}{ccc}\hfill 0.05\left(600\right)+34& =& 64\hfill \\ \hfill 0.04\left(600\right)+40& =& 64\hfill \end{array}$

Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See Figure 2

Try It

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?

$C=2.5x+3,650$

Access these online resources for additional instruction and practice with linear equations.

Key Concepts

• We can solve linear equations in one variable in the form $\,ax+b=0\,$ using standard algebraic properties. See Example 1  and Example 2.
• A linear equation can be used to solve for an unknown in a number problem. See Example 3.
• Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities. See Example 4.

Glossary

conditional equation
an equation that is true for some values of the variable
identity equation
an equation that is true for all values of the variable
inconsistent equation
an equation producing a false result
linear equation
an algebraic equation in which each term is either a constant or the product of a constant and the first power of a variable
solution set
the set of all solutions to an equation
slope
the change in y-values over the change in x-values