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Section 1.1 Linear Equations in One Variable

Before Starting this Section, Review

1 Algebraic expressions (Section P.1 , page 14)
2 Like terms (Section P.3 , page 31)
3 Arithmetic of fractions, least common denominator (Section P.5 , page 54)

Objectives

1 Learn the vocabulary and concepts used in studying equations.
2 Solve linear equations in one variable.
3 Solve formulas for a specific variable.

Bungee-Jumping TV Contestants

England’s Oxford Dangerous Sports Club started the modern version of the sport of “bungee” (or bungy) jumping on April 1, 1978, from the Clifton Suspension Bridge in Bristol, England. Bungee cords consist of hundreds of continuous-length rubber strands encased in a nylon sheath. One end of the cord is tied to a solid structure such as a crane or bridge, and the other end is tied via padded ankle straps to the jumper’s ankle. The jumper jumps, and if things go as planned, the cord extends like a rubber band and the jumper bounces up and down until coming to a halt. Suppose a television adventure program has its contestants bungee jump from a bridge 120 feet above the water. The bungee cord that is used has a 140%–150% elongation, meaning that its extended length will be the original length plus a maximum of an additional 150% of its original length. If the show’s producer wants to make sure the jumper does not get closer than 10 feet to the water and if no contestant is more than 7 feet tall, how long can the bungee cord be? Using the information from this section, we learn in Example 6 that the cord should be about 41 feet long.

Definitions

1 Learn the vocabulary and concepts used in studying equations.

An equation is a statement that two mathematical expressions are equal. For example,

7 - 5 = 2

$7 - 5 = 2$

is an equation. In algebra, however, we are generally more interested in equations that contain variables.

An equation in one variable is a statement that two expressions, with at least one containing the variable, are equal. For example, $2 x - 3 = 7$ $2 x - 3 = 7$ is an equation in the variable x The expressions $2 x - 3$ $2 x - 3$ and 7 are the sides of the equation. The domain of the variable in an equation is the set of all real numbers for which both sides of the equation are defined.

Side Note

When finding the domain of a variable, remember that

Division by 0 is undefined.
The square root (or any even root) of a negative number is not a real number.

Example 1 Finding the Domain of the Variable

Find the domain of the variable x in each of the following equations.

$2 = \frac{5}{x - 1}$ $2 = \frac{5}{x - 1}$
$x - 2 = \sqrt{x}$ $x - 2 = \sqrt{x}$
$2 x - 3 = 7$ $2 x - 3 = 7$

Solution

In the equation $2 = \frac{5}{x - 1},$ $2 = \frac{5}{x - 1},$ the left side of the equation does not contain x. The right side of the equation, $\frac{5}{x - 1}$ $\frac{5}{x - 1}$ , is not defined if $x = 1$ $x = 1$ because division by 0 is undefined. The domain of x is the set of all real numbers except the number 1.
The square root of a negative number is not a real number; so $\sqrt{x},$ $\sqrt{x},$ the right side of the equation $x - 2 = \sqrt{x},$ $x - 2 = \sqrt{x},$ is defined only when $x \geq 0 .$ $x \geq 0 .$ The left side is defined for all real numbers. So the domain of x in this equation is ${x | x \geq 0},$ ${x | x \geq 0},$ or in interval notation, $[0, \infty) .$ $[0, \infty) .$
Because both sides of the equation $2 x - 3 = 7$ $2 x - 3 = 7$ are defined for all real numbers, the domain is ${x | x is a real number},$ ${x | x is a real number},$ or in interval notation, $(- \infty, \infty) .$ $(- \infty, \infty) .$

Practice Problem 1

Find the domain of the variable x in each of the following equations.
1. $\frac{x}{3} - 7 = 5$ $\frac{x}{3} - 7 = 5$
2. $\frac{3}{2 - x} = 4$ $\frac{3}{2 - x} = 4$
3. $\sqrt{x - 1} = 0$ $\sqrt{x - 1} = 0$

Side Note

Frequently, when only a few numbers are excluded from the domain of the variable, we write the exceptions on the side of the equation. In Example 1 part a, if we wanted to specify the domain of the variable, we would simply write $2 = \frac{5}{x - 1}, x \neq 1 .$ $2 = \frac{5}{x - 1}, x \neq 1 .$ This domain can be written in interval notation as $(- \infty, 1) \cup (1, \infty) .$ $(- \infty, 1) \cup (1, \infty) .$

When the variable in an equation is replaced with a specific value from its domain, the resulting statement may be true or false. For example, in the equation $2 x - 3 = 7,$ $2 x - 3 = 7,$ if we let $x = 1,$ $x = 1,$ we obtain

2 (1) - 3 = 7 or - 1 = 7,

$2 (1) - 3 = 7 or - 1 = 7,$

which is obviously false. However, if we replace x with 5, we obtain $2 (5) - 3 = 7,$ $2 (5) - 3 = 7,$ a true statement. Those values (if any) of the variable that result in a true statement are called solutions or roots of the equation. Thus, 5 is a solution (or root) of the equation $2 x - 3 = 7 .$ $2 x - 3 = 7 .$ We also say that 5 satisfies the equation $2 x - 3 = 7 .$ $2 x - 3 = 7 .$ To solve an equation means to find all solutions of the equation; the set of all solutions of an equation is called its solution set. The process of finding the solution set of an equation is called solving the equation.

Identities, Conditional Equations, and Inconsistent Equations

An equation that is satisfied by every real number in the domain of the variable is called an identity. The equations

2 (x + 3) = 2 x + 6 and x^{2} - 9 = (x + 3) (x - 3)

$2 (x + 3) = 2 x + 6 and x^{2} - 9 = (x + 3) (x - 3)$

are examples of identities. You should recognize that these equations are examples of the distributive property and the difference of squares, respectively.

An equation that is not an identity but is satisfied by at least one real number is called a conditional equation. To verify that an equation is a conditional equation, you must find at least one number that is a solution of the equation and at least one number that is not a solution of the equation. The equation $2 x - 3 = 7$ $2 x - 3 = 7$ is an example of a conditional equation: 5 satisfies the equation, but 1 does not.

An equation that no number satisfies is called an inconsistent equation. The solution set of an inconsistent equation is designated by the empty set symbol, $\emptyset .$ $\emptyset .$ The equation $x = x + 5$ $x = x + 5$ is an example of an inconsistent equation. It is inconsistent because no number is 5 more than itself.

In summary, there are three types of equations: identities, conditional equations, and inconsistent equations.

Equivalent Equations

Equations that have the same solution set are called equivalent equations. For example, equations $x = 4$ $x = 4$ and $3 x = 12$ $3 x = 12$ are equivalent equations with solution set {4}.

Side Note

Squaring both sides of an equation does not result in an equivalent equation. For example, $x = 2$ $x = 2$ has one solution but $x^{2} = 4$ $x^{2} = 4$ has two solutions, $x = 2$ $x = 2$ and $x = - 2 .$ $x = - 2 .$

In general, to solve an equation in one variable, we replace the given equation with a sequence of equivalent equations until we obtain an equation whose solution is obvious, such as $x = 4 .$ $x = 4 .$ The following operations yield equivalent equations.

Procedures That Result in Equivalent Equations
Procedures	Given Equation	Equivalent Equation
Simplify expressions on either side by eliminating parentheses, combining like terms, etc.	$(2 x - 1) - (x + 1) = 4$ $(2 x - 1) - (x + 1) = 4$	$\begin{array}{l} 2 x & - & 1 & - & x - 1 & = & 4 \\ or & x - 2 & = & 4 \end{array}$ $\begin{array}{l} 2 x & - & 1 & - & x - 1 & = & 4 \\ or & x - 2 & = & 4 \end{array}$
Add (or subtract) the same expression (with the same domain) on both sides of the equation.	$x - 2 = 4$ $x - 2 = 4$	$\begin{array}{l} x & - & 2 & + & 2 & = & 4 + 2 \\ or & x & = & 6 \end{array}$ $\begin{array}{l} x & - & 2 & + & 2 & = & 4 + 2 \\ or & x & = & 6 \end{array}$
Multiply (or divide) both sides of the equation by the same nonzero number.	$2 x = 8$ $2 x = 8$	$\begin{array}{lrcl} \frac{1}{2} \cdot & 2 x & = & \frac{1}{2} \cdot 8 \\ or & x & = & 4 \end{array}$ $\begin{array}{lrcl} \frac{1}{2} \cdot & 2 x & = & \frac{1}{2} \cdot 8 \\ or & x & = & 4 \end{array}$
Interchange the two sides of the equation.	$- 3 = x$ $- 3 = x$	$x = - 3$ $x = - 3$

Solving Linear Equations in One Variable

2 Solve linear equations in one variable.

Because the highest exponent on the variable in a linear equation is $1 (x = x^{1}),$ $1 (x = x^{1}),$ a linear equation is also called a first-degree equation. We can see that the linear equation $a x + b = 0$ $a x + b = 0$ with $a \neq 0$ $a \neq 0$ has exactly one solution by using the following steps:

\begin{array}{rcll} a x + b & = & 0 & Original equation, a \neq 0 \\ a x & = & - b & Subtract b from both sides . \\ x & = & - \frac{b}{a} & Divide both sides by a . \end{array}

$\begin{array}{rcll} a x + b & = & 0 & Original equation, a \neq 0 \\ a x & = & - b & Subtract b from both sides . \\ x & = & - \frac{b}{a} & Divide both sides by a . \end{array}$

In the Procedure in Action box, we describe a step-by-step procedure for solving a linear equation in one variable.

Procedure in Action

Example 2 Solving Conditional Linear Equations in One Variable

OBJECTIVE	EXAMPLE
Solve a linear equation in one variable.	*Solve* $\frac{1}{3} x - \frac{2}{3} = \frac{1}{2} (1 - 3 x) .$ $\frac{1}{3} x - \frac{2}{3} = \frac{1}{2} (1 - 3 x) .$
Step 1 Eliminate fractions, if any. Multiply both sides of the equation by the least common denominator (LCD) of all of the fractions.	$\begin{array}{rcll} \frac{1}{3} x - \frac{2}{3} & = & \frac{1}{2} (1 - 3 x) & Original equation \\ 6 (\frac{1}{3} x - \frac{2}{3}) & = & 6 \cdot \frac{1}{2} (1 - 3 x) & \begin{array}{l} Multiply both sides by \\ 6 (LCD) to eliminate \\ fractions . \end{array} \end{array}$ $\begin{array}{rcll} \frac{1}{3} x - \frac{2}{3} & = & \frac{1}{2} (1 - 3 x) & Original equation \\ 6 (\frac{1}{3} x - \frac{2}{3}) & = & 6 \cdot \frac{1}{2} (1 - 3 x) & \begin{array}{l} Multiply both sides by \\ 6 (LCD) to eliminate \\ fractions . \end{array} \end{array}$
Step 2 Simplify. Simplify both sides of the equation by removing parentheses and other grouping symbols (if any) and combining like terms.	$\begin{array}{rcll} 6 \cdot \frac{1}{3} x - 6 \cdot \frac{2}{3} & = & 3 (1 - 3 x) & \begin{array}{l} Distribute on the left side; \\ simplify on the right . \end{array} \\ 2 x - 4 & = & 3 - 9 x & \begin{array}{l} Distribute on the left side; \\ simplify on the right . \end{array} \end{array}$ $\begin{array}{rcll} 6 \cdot \frac{1}{3} x - 6 \cdot \frac{2}{3} & = & 3 (1 - 3 x) & \begin{array}{l} Distribute on the left side; \\ simplify on the right . \end{array} \\ 2 x - 4 & = & 3 - 9 x & \begin{array}{l} Distribute on the left side; \\ simplify on the right . \end{array} \end{array}$
Step 3 Isolate the variable term. Add appropriate expressions to both sides so that when both sides are simplified, the terms containing the variable are on one side and all constant terms are on the other side.	$\begin{array}{l} 2 x - 4 + 9 x + 4 = 3 - 9 x + 9 x + 4 & Add 9 x + 4 \\ to both sides . \end{array}$ $\begin{array}{l} 2 x - 4 + 9 x + 4 = 3 - 9 x + 9 x + 4 & Add 9 x + 4 \\ to both sides . \end{array}$
Step 4 Combine terms. Combine terms containing the variable to obtain one term that contains the variable as a factor.	$\begin{array}{l} 11 x = 7 & Collect like terms and \\ combine constants . \end{array}$ $\begin{array}{l} 11 x = 7 & Collect like terms and \\ combine constants . \end{array}$
Step 5 Isolate the variable. Divide both sides by the coefficient of the variable to obtain the solution.	$\begin{array}{l} x = \frac{7}{11} & Divide both sides by 11 . \end{array}$ $\begin{array}{l} x = \frac{7}{11} & Divide both sides by 11 . \end{array}$
Step 6 Check the solution. Substitute the solution into the original equation to check for computational errors.	Check: Substitute $\frac{7}{11}$ $\frac{7}{11}$ for `x` in both sides of the original equation. The result should be $- \frac{5}{11} = - \frac{5}{11} .$ $- \frac{5}{11} = - \frac{5}{11} .$

Practice Problem 2

Solve $\frac{2}{3} - \frac{3}{2} x = \frac{1}{6} - \frac{7}{3} x .$ $\frac{2}{3} - \frac{3}{2} x = \frac{1}{6} - \frac{7}{3} x .$

Example 3 Solving a Linear Equation

Solve $6 x - [3 x - 2 (x - 2)] = 11 .$ $6 x - [3 x - 2 (x - 2)] = 11 .$

Solution

Step 2 $\begin{array}{rcll} 6 x - [3 x - 2 (x - 2)] & = & 11 & Original equation \\ 6 x - [3 x - 2 x + 4] & = & 11 & Remove innermost parentheses by \\ distributing - 2 (x - 2) = - 2 x + 4 . \\ 6 x - 3 x + 2 x - 4 & = & 11 & Remove square brackets and change \\ the sign of each enclosed term . \\ 5 x - 4 & = & 11 & Combine like terms . \end{array}$ $\begin{array}{rcll} 6 x - [3 x - 2 (x - 2)] & = & 11 & Original equation \\ 6 x - [3 x - 2 x + 4] & = & 11 & Remove innermost parentheses by \\ distributing - 2 (x - 2) = - 2 x + 4 . \\ 6 x - 3 x + 2 x - 4 & = & 11 & Remove square brackets and change \\ the sign of each enclosed term . \\ 5 x - 4 & = & 11 & Combine like terms . \end{array}$
Step 3 $\begin{matrix} 5 x - 4 + 4 = 11 + 4 & Add 4 to both sides . \end{matrix}$ $\begin{matrix} 5 x - 4 + 4 = 11 + 4 & Add 4 to both sides . \end{matrix}$
Step 4 $\begin{matrix} 5 x = 15 & Simplify both sides in Step 3 . \end{matrix}$ $\begin{matrix} 5 x = 15 & Simplify both sides in Step 3 . \end{matrix}$
Step 5 $\begin{aligned} \frac{5 x}{5} & = & \frac{15}{5} & Divide both sides by 5 . \\ x & = & 3 & Simplify . \end{aligned}$ $\begin{aligned} \frac{5 x}{5} & = & \frac{15}{5} & Divide both sides by 5 . \\ x & = & 3 & Simplify . \end{aligned}$

The apparent solution is 3.
Step 6 Check: Substitute $x = 3$ $x = 3$ into the original equation. You should obtain $11 = 11 .$ $11 = 11 .$ Thus, 3 is the only solution of the original equation; so {3} is its solution set.

Practice Problem 3

Solve $3 x - [2 x - 6 (x + 1)] = - 1 .$ $3 x - [2 x - 6 (x + 1)] = - 1 .$

If solving a linear equation results in a false statement, such as $4 = - 3,$ $4 = - 3,$ the equation is inconsistent.

Example 4 Solving a Linear Equation That Is Inconsistent

Solve $5 x - [3 x + 2 (x - 1)] = 1 .$ $5 x - [3 x + 2 (x - 1)] = 1 .$

Solution

Step 2 $\begin{array}{rcll} 5 x - [3 x + 2 (x - 1)] & = & 1 & Original equation \\ 5 x - [3 x + 2 x - 2] & = & 1 & Remove innermost parentheses by \\ distributing 2 (x - 1) = 2 x - 2 . \\ 5 x - 3 x - 2 x + 2 & = & 1 & Remove square brackets and change \\ the sign of each enclosed term . \\ 2 & = & 1 & Combine like terms . \end{array}$ $\begin{array}{rcll} 5 x - [3 x + 2 (x - 1)] & = & 1 & Original equation \\ 5 x - [3 x + 2 x - 2] & = & 1 & Remove innermost parentheses by \\ distributing 2 (x - 1) = 2 x - 2 . \\ 5 x - 3 x - 2 x + 2 & = & 1 & Remove square brackets and change \\ the sign of each enclosed term . \\ 2 & = & 1 & Combine like terms . \end{array}$

Because $2 = 1$ $2 = 1$ is false, no number satisfies this equation. It is inconsistent.

Practice Problem 4

Solve $3 x - [2 x - 6 (x + 1)] = 7 x - 1 .$ $3 x - [2 x - 6 (x + 1)] = 7 x - 1 .$

If solving a linear equation results in an equivalent equation that is always true, such as $3 = 3,$ $3 = 3,$ the equation is an identity.

Example 5 Solving an Equation That Is an Identity

Solve $1 - 5 y + 2 (y + 7) = 2 y + 5 (3 - y) .$ $1 - 5 y + 2 (y + 7) = 2 y + 5 (3 - y) .$

Solution

Step 2 $\begin{array}{rcll} 1 - 5 y + 2 (y + 7) & = & 2 y + 5 (3 - y) & Original equation \\ 1 - 5 y + 2 y + 14 & = & 2 y + 15 - 5 y & Distributive property \\ 15 - 3 y & = & 15 - 3 y & Collect like terms and \\ combine constants on each side . \end{array}$ $\begin{array}{rcll} 1 - 5 y + 2 (y + 7) & = & 2 y + 5 (3 - y) & Original equation \\ 1 - 5 y + 2 y + 14 & = & 2 y + 15 - 5 y & Distributive property \\ 15 - 3 y & = & 15 - 3 y & Collect like terms and \\ combine constants on each side . \end{array}$
Step 3 $\begin{array}{rcll} 15 - 3 y + 3 y & = & 15 - 3 y + 3 y & Add 3 y to both sides . \\ 15 & = & 15 & Simplify . \end{array}$ $\begin{array}{rcll} 15 - 3 y + 3 y & = & 15 - 3 y + 3 y & Add 3 y to both sides . \\ 15 & = & 15 & Simplify . \end{array}$

We have shown that $15 = 15$ $15 = 15$ , a true statement, is equivalent to the original equation. Consequently, the original equation is an identity and the solution set is $(- \infty, \infty) .$ $(- \infty, \infty) .$

Practice Problem 5

Solve the equation $2 (3 x - 6) + 5 = 12 - (19 - 6 x) .$ $2 (3 x - 6) + 5 = 12 - (19 - 6 x) .$

Identifying Types of Linear Equations

An equation that is satisfied by all values in the domain of the variable is an identity. For example, $2 (x - 1) = 2 x - 2 .$ $2 (x - 1) = 2 x - 2 .$ When you try to solve an identity, you get a true statement such as $3 = 3$ $3 = 3$ or $0 = 0 .$ $0 = 0 .$
An equation that is not an identity but is satisfied by at least one number in the domain of the variable is a conditional equation. For example, $2 x = 6$ $2 x = 6$ is a linear conditional equation; $x = 3$ $x = 3$ is a solution, but $x = 0$ $x = 0$ is not.
An equation that is not satisfied by any value of the variable is an inconsistent equation. For example, $x = x + 2$ $x = x + 2$ is an inconsistent equation. Because no number is 2 more than itself, the solution set of the equation $x = x + 2$ $x = x + 2$ is $\emptyset .$ $\emptyset .$ When you try to solve an inconsistent equation, you obtain a false statement such as $0 = 2 .$ $0 = 2 .$

Example 6 Bungee-Jumping TV Contestants

The introduction to this section described a television show on which a bungee cord with 140%–150% elongation is used for contestants’ bungee jumping from a bridge 120 feet above the water. The show’s producer wants to make sure the jumper does not get closer than 10 feet to the water. How long can the bungee cord be, given that no contestant is more than 7 feet tall?

Solution

Step 1 We want the extended cord plus the length of the jumper’s body to be a minimum of 10 feet above the water. To be safe, we determine the maximum extended length of the cord, add the height of the tallest possible contestant, and ensure that this total length is 10 feet above the water.
Step 2 Let $x = length,$ $x = length,$ in feet, of the cord to be used.
Step 3 Then $x + 1.5 x = 2.5 x$ $x + 1.5 x = 2.5 x$ is the maximum extended length (in feet) of the cord (with 150% elongation). We have

$\begin{array}{l} \begin{array}{l} Extended \\ cord length \end{array} + \begin{array}{l} Body \\ length \end{array} + \begin{array}{l} 10 -foot \\ buffer \end{array} = \begin{array}{l} Height of bridge \\ above the water \end{array} \\ \begin{array}{rcll} 2.5 x + 7 + 10 & = & 120 & \begin{array}{l} Replace descriptions with \\ arithmetic expressions . \end{array} \end{array} \end{array}$ $\begin{array}{l} \begin{array}{l} Extended \\ cord length \end{array} + \begin{array}{l} Body \\ length \end{array} + \begin{array}{l} 10 -foot \\ buffer \end{array} = \begin{array}{l} Height of bridge \\ above the water \end{array} \\ \begin{array}{rcll} 2.5 x + 7 + 10 & = & 120 & \begin{array}{l} Replace descriptions with \\ arithmetic expressions . \end{array} \end{array} \end{array}$
Step 4

$\begin{array}{rcll} 2.5 x + 17 & = & 120 & Combine constants . \\ 2.5 x & = & 103 & Subtract 17 from both sides . \\ x & = & 41.2 feet & Divide both sides by 2.5 . \end{array}$ $\begin{array}{rcll} 2.5 x + 17 & = & 120 & Combine constants . \\ 2.5 x & = & 103 & Subtract 17 from both sides . \\ x & = & 41.2 feet & Divide both sides by 2.5 . \end{array}$
Step 5 If the length of the cord is less than 41.2 feet, all conditions are met.
Step 6 Check: $41.2 + 1.5 (41.2) + 7 + 10 = 120;$ $41.2 + 1.5 (41.2) + 7 + 10 = 120;$ so any length less than 41.2 ft will work.

Practice Problem 6

What length of cord should be used in Example 6 if the elongation is 200% instead of 140%–150%?

Formulas

3 Solve formulas for a specific variable.

Solving equations that arise from everyday experiences is one of the most important uses of algebra. An equation that expresses a relationship between two or more variables is called a formula.

Example 7 Converting Temperatures

The formula for converting the temperature in degrees Celsius (C) to degrees Fahrenheit (F) is

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

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

While planning a trip to the United States, a London resident notes that the local temperature will be $86 °$ $86 °$ Fahrenheit. What is the temperature in degrees Celsius? (Temperatures are recorded in Celsius degrees in England.)

Solution

We first substitute 86 for F in the formula

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

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

to obtain

86 = \frac{9}{5} C + 32 .

$86 = \frac{9}{5} C + 32 .$

We now solve this equation for C

\begin{array}{rcll} 5 (86) & = & 5 (\frac{9}{5} C + 32) & Multiply both sides by 5 . \\ 430 & = & 9 C + 160 & Simplify . \\ 270 & = & 9 C & Subtract 160 from both sides . \\ \frac{270}{9} & = & C & Divide both sides by 9 . \\ 30 & = & C & Simplify . \end{array}

$\begin{array}{rcll} 5 (86) & = & 5 (\frac{9}{5} C + 32) & Multiply both sides by 5 . \\ 430 & = & 9 C + 160 & Simplify . \\ 270 & = & 9 C & Subtract 160 from both sides . \\ \frac{270}{9} & = & C & Divide both sides by 9 . \\ 30 & = & C & Simplify . \end{array}$

Thus, $86 ° F$ $86 ° F$ converts to $30 ° C .$ $30 ° C .$

Practice Problem 7

Use the formula in Example 7 to find the temperature in degrees Celsius, assuming that the temperature is $50 °$ $50 °$ Fahrenheit.

In Example 7, we specified $F = 86$ $F = 86$ and solved the equation for C. A more general result can be obtained by assuming that F is a fixed number, or constant, and solving for C. This process is called solving for a specified variable.

Example 8 Solving for a Specified Variable

Solve $F = \frac{9}{5} C + 32$ $F = \frac{9}{5} C + 32$ for C.
Solve $S = 2 ℓ w + 2 ℓ h + 2 w h$ $S = 2 ℓ w + 2 ℓ h + 2 w h$ for h.

Solution

$\begin{array}{l} \begin{array}{l} F = \frac{9}{5} C + 32 & Original equation \end{array} \\ \begin{array}{rcll} F - 32 & = & \frac{9}{5} C & Subtract 32 from both sides . \\ \frac{5}{9} (F - 32) & = & C & Multiply both sides by \frac{5}{9}; \frac{5}{9} \cdot \frac{9}{5} = 1 . \\ or C & = & \frac{5}{9} (F - 32) \end{array} \end{array}$ $\begin{array}{l} \begin{array}{l} F = \frac{9}{5} C + 32 & Original equation \end{array} \\ \begin{array}{rcll} F - 32 & = & \frac{9}{5} C & Subtract 32 from both sides . \\ \frac{5}{9} (F - 32) & = & C & Multiply both sides by \frac{5}{9}; \frac{5}{9} \cdot \frac{9}{5} = 1 . \\ or C & = & \frac{5}{9} (F - 32) \end{array} \end{array}$
$\begin{array}{rcll} S = 2 ℓ w & + & 2 ℓ h + 2 w h for h & Original equation \\ S - 2 ℓ w & = & 2 ℓ h + 2 w h & \begin{array}{l} Subtract 2 ℓ w from both sides so that \\ all terms with h are on one side . \end{array} \\ S - 2 ℓ w & = & (2 ℓ + 2 w) h & Factor out h . \\ \frac{S - 2 ℓ w}{2 ℓ + 2 w} & = & h & Divide both sides by 2 ℓ + 2 w . \\ or h & = & \frac{S - 2 ℓ w}{2 ℓ + 2 w} \end{array}$ $\begin{array}{rcll} S = 2 ℓ w & + & 2 ℓ h + 2 w h for h & Original equation \\ S - 2 ℓ w & = & 2 ℓ h + 2 w h & \begin{array}{l} Subtract 2 ℓ w from both sides so that \\ all terms with h are on one side . \end{array} \\ S - 2 ℓ w & = & (2 ℓ + 2 w) h & Factor out h . \\ \frac{S - 2 ℓ w}{2 ℓ + 2 w} & = & h & Divide both sides by 2 ℓ + 2 w . \\ or h & = & \frac{S - 2 ℓ w}{2 ℓ + 2 w} \end{array}$

Practice Problem 8

Solve $P = 2 ℓ + 2 w$ $P = 2 ℓ + 2 w$ for w.

Example 9 Applying a Geometry Formula

A cereal box that is 12 inches high and 8 inches wide holds 240 cubic inches of cereal. How deep is the box?

Solution

Use the following formula: $V = h w d,$ $V = h w d,$ where V is the volume, h is the height, w is the width, and d is the depth of the cereal box.

\begin{array}{rcll} 240 & = & (12) (8) d & Replace V with 240, h with 12, and w with 8 . \\ d & = & 2.5 & Divide both sides by 96 . \end{array}

$\begin{array}{rcll} 240 & = & (12) (8) d & Replace V with 240, h with 12, and w with 8 . \\ d & = & 2.5 & Divide both sides by 96 . \end{array}$

The cereal box is 2.5 inches deep.

Practice Problem 9

A rectangular suitcase is 21 inches high, 13 inches wide, and 9 inches deep. Find the volume of the suitcase.

Section 1.1 Exercises

Concepts and Vocabulary

The domain of the variable in an equation is the set of all real numbers for which both sides of the equation are .
Standard form for a linear equation in x is .
Two equations with the same solution set are called .
A conditional equation is an equation that is not true for values of the variables.
True or False. A linear equation that is equivalent to the equation $0 = 4$ $0 = 4$ is a conditional equation.
True or False. A linear equation that is equivalent to the equation $- 1 = - 1$ $- 1 = - 1$ is an identity.
True or False. The equation $x - 2 = 0$ $x - 2 = 0$ is equivalent to the equation $6 - 3 x = 0 .$ $6 - 3 x = 0 .$
True or False. Solving $P + 2 b = ℓ$ $P + 2 b = ℓ$ for b results in $b = \frac{ℓ}{2} - P .$ $b = \frac{ℓ}{2} - P .$

Building Skills

In Exercises 9–13, determine whether the given value of the variable is a solution of the equation.

$x - 2 = 5 x + 6$ $x - 2 = 5 x + 6$
1. $x = 0$ $x = 0$
2. $x = - 2$ $x = - 2$
$8 x + 3 = 14 x - 1$ $8 x + 3 = 14 x - 1$
1. $x = - 1$ $x = - 1$
2. $x = \frac{2}{3}$ $x = \frac{2}{3}$
$\frac{2}{x} = \frac{1}{3} + \frac{1}{x + 2}$ $\frac{2}{x} = \frac{1}{3} + \frac{1}{x + 2}$
1. $x = 4$ $x = 4$
2. $x = 1$ $x = 1$
$(x - 3) (2 x + 1) = 0$ $(x - 3) (2 x + 1) = 0$
1. $x = \frac{1}{2}$ $x = \frac{1}{2}$
2. $x = 3$ $x = 3$
$2 x + 3 x = 5 x$ $2 x + 3 x = 5 x$
1. $x = 157$ $x = 157$
2. $x = - 2046$ $x = - 2046$

In Exercises 14–18, find the domain of the variable in each equation. Write the answer in interval notation.

$(2 - x) - 4 x = 7 - 3 (x + 4)$ $(2 - x) - 4 x = 7 - 3 (x + 4)$
$\frac{y}{y - 1} = \frac{3}{y + 2}$ $\frac{y}{y - 1} = \frac{3}{y + 2}$
$\frac{1}{y} = 2 + \sqrt{y}$ $\frac{1}{y} = 2 + \sqrt{y}$
$\frac{3 x}{(x - 3) (x - 4)} = 2 x + 9$ $\frac{3 x}{(x - 3) (x - 4)} = 2 x + 9$
$\frac{1}{\sqrt{x}} = x^{2} - 1$ $\frac{1}{\sqrt{x}} = x^{2} - 1$

In Exercises 19–22, determine whether the given equation is an identity. If the equation is not an identity, give a value that demonstrates that fact.

$2 x + 3 = 5 x + 1$ $2 x + 3 = 5 x + 1$
$3 x + 4 = 6 x + 2 - (3 x - 2)$ $3 x + 4 = 6 x + 2 - (3 x - 2)$
$\frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2 x}$ $\frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2 x}$
$\frac{1}{x + 3} = \frac{1}{x} + \frac{1}{3}$ $\frac{1}{x + 3} = \frac{1}{x} + \frac{1}{3}$

In Exercises 23–48, solve each equation.

$3 x + 5 = 14$ $3 x + 5 = 14$
$2 x - 17 = 7$ $2 x - 17 = 7$
$- 10 x + 12 = 32$ $- 10 x + 12 = 32$
$- 2 x + 5 = 6$ $- 2 x + 5 = 6$
$3 - y = - 4$ $3 - y = - 4$
$2 - 7 y = 23$ $2 - 7 y = 23$
$7 x + 7 = 2 (x + 1)$ $7 x + 7 = 2 (x + 1)$
$3 (x + 2) = 4 - x$ $3 (x + 2) = 4 - x$
$3 (2 - y) + 5 y = 3 y$ $3 (2 - y) + 5 y = 3 y$
$9 y - 3 (y - 1) = 6 + y$ $9 y - 3 (y - 1) = 6 + y$
$4 y - 3 y + 7 - y = 2 - (7 - y)$ $4 y - 3 y + 7 - y = 2 - (7 - y)$
$3 (y - 1) = 6 y - 4 + 2 y - 4 y$ $3 (y - 1) = 6 y - 4 + 2 y - 4 y$
$3 (x - 2) + 2 (3 - x) = 1$ $3 (x - 2) + 2 (3 - x) = 1$
$2 x - 3 - (3 x - 1) = 6$ $2 x - 3 - (3 x - 1) = 6$
$2 x + 3 (x - 4) = 7 x + 10$ $2 x + 3 (x - 4) = 7 x + 10$
$3 (2 - 3 x) - 4 x = 3 x - 10$ $3 (2 - 3 x) - 4 x = 3 x - 10$
$4 [x + 2 (3 - x)] = 2 x + 1$ $4 [x + 2 (3 - x)] = 2 x + 1$
$3 - [x - 3 (x + 2)] = 4$ $3 - [x - 3 (x + 2)] = 4$
$3 (4 y - 3) = 4 [y - (4 y - 3)]$ $3 (4 y - 3) = 4 [y - (4 y - 3)]$
$5 - (6 y + 9) + 2 y = 2 (y + 1)$ $5 - (6 y + 9) + 2 y = 2 (y + 1)$
$2 x - 3 (2 - x) = (x - 3) + 2 x + 1$ $2 x - 3 (2 - x) = (x - 3) + 2 x + 1$
$5 (x - 3) - 6 (x - 4) = - 5$ $5 (x - 3) - 6 (x - 4) = - 5$
$\frac{2 x + 1}{9} - \frac{x + 4}{6} = 1$ $\frac{2 x + 1}{9} - \frac{x + 4}{6} = 1$
$\frac{2 - 3 x}{7} + \frac{x - 1}{3} = \frac{3 x}{7}$ $\frac{2 - 3 x}{7} + \frac{x - 1}{3} = \frac{3 x}{7}$
$\frac{1 - x}{4} + \frac{5 x + 1}{2} = 3 - \frac{2 (x + 1)}{8}$ $\frac{1 - x}{4} + \frac{5 x + 1}{2} = 3 - \frac{2 (x + 1)}{8}$
$\frac{x + 4}{3} + 2 x - \frac{1}{2} = \frac{3 x + 2}{6}$ $\frac{x + 4}{3} + 2 x - \frac{1}{2} = \frac{3 x + 2}{6}$

In Exercises 49–60, solve each formula for the indicated variable.

$d = r t$ $d = r t$ for r.
$F = m a$ $F = m a$ for a.
$C = 2 π r$ $C = 2 π r$ for r
$A = 2 π r x + π r^{2}$ $A = 2 π r x + π r^{2}$ for x
$I = \frac{E}{R}$ $I = \frac{E}{R}$ for R
$A = P (1 + r t)$ $A = P (1 + r t)$ for t
$A = \frac{(a + b) h}{2}$ $A = \frac{(a + b) h}{2}$ for h
$T = a + (n - 1) d$ $T = a + (n - 1) d$ for d
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ for u
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ for $R_{2}$ $R_{2}$
$y = m x + b$ $y = m x + b$ for m
$a x + b y = c$ $a x + b y = c$ for y

Applying the Concepts

Swimming pool dimensions. The volume of a swimming pool is 2808 cubic feet. Assuming that the pool is 18 feet long and 12 feet deep, find its width.
Hole dimensions. If a box-shaped hole 7 feet long and 3 feet wide holds 168 cubic feet of dirt, how deep must the hole be?
Geometry. The perimeter of a rectangle is 80 feet. Find its dimensions, assuming that its length is 5 feet less than twice its width.
Geometry. The width of a rectangle is 3 feet more than one-half its length, and the perimeter of the rectangle is 36 feet. Find its dimensions.
Geometry. A circle has a circumference of $114 π$ $114 π$ centimeters. Find the radius of the circle.
Geometry. A rectangle has a perimeter of 28 meters and a width of 5 meters. Find the length of the rectangle.
Storage barrel dimensions. A barrel has a surface area of $6 π$ $6 π$ square meters and a radius of 1 meter. Find the height of the barrel.
Can dimensions. A can has a volume of $148 π$ $148 π$ cubic centimeters and a radius of 2 centimeters. Find the height of the can.
Geometry. A trapezoid has an area of 66 square feet and a height of 6 feet. If one base is 3 feet, what is the length of the other base?
Geometry. A trapezoid has an area of 35 square centimeters. Assuming that one base is 9 centimeters and the other base is 11 centimeters, find the height of the trapezoid.
Temperature on Mars. On summer days on Mars, close to its equator, the temperature can be around $20 °$ $20 °$ Celsius. What is this temperature in degrees Fahrenheit?

(Source: www.universetoday.com)
Temperature on the moon. The average daytime temperature on the bright side of the moon is $224.6 °$ $224.6 °$ Fahrenheit. What is this temperature in degrees Celsius?

(Source: www.space.com)
Temperature in Honolulu, Hawaii. On a rare cold day in Honolulu, the temperature reached $57 °$ $57 °$ Fahrenheit. What is this temperature in degrees Celsius, rounded to the nearest degree?
Temperature in Las Vegas, Nevada. On an unusually hot day in Las Vegas, the temperature reached $47.8 °$ $47.8 °$ Celsius. What is this temperature in degrees Fahrenheit, rounded to the nearest degree?

Beyond the Basics

Find a value of k so that $3 x - 1 = k$ $3 x - 1 = k$ and $7 x + 2 = 16$ $7 x + 2 = 16$ are equivalent.
Find a value of k so that the equation $\frac{5}{y - 4} = \frac{6}{y + k}$ $\frac{5}{y - 4} = \frac{6}{y + k}$ has solution set {9}.
Find a value of k so that the equation $\frac{3}{y - 2} = \frac{4}{y + k}$ $\frac{3}{y - 2} = \frac{4}{y + k}$ is inconsistent.
Find a value of k so that the equation $\frac{1}{x^{2} - 9} = \frac{1}{(x - 3) (x + k)}$ $\frac{1}{x^{2} - 9} = \frac{1}{(x - 3) (x + k)}$ is an identity.

In Exercises 79–84, solve each equation for x. Assume $a \neq 0$ $a \neq 0$ and $b \neq 0 .$ $b \neq 0 .$

$a (a + x) = b^{2} - b x$ $a (a + x) = b^{2} - b x$
$9 + a^{2} x - a x = 6 x + a^{2}$ $9 + a^{2} x - a x = 6 x + a^{2}$
$\frac{a x}{b} - \frac{b x}{a} = \frac{{(a + b)}^{2}}{a b}$ $\frac{a x}{b} - \frac{b x}{a} = \frac{{(a + b)}^{2}}{a b}$
$\frac{2}{3} - \frac{x}{3 b} - \frac{2 x + b}{2 a} + \frac{6 b^{2} + a^{2}}{6 a b} = 0$ $\frac{2}{3} - \frac{x}{3 b} - \frac{2 x + b}{2 a} + \frac{6 b^{2} + a^{2}}{6 a b} = 0$
$\frac{b (b x - 1)}{a} - \frac{a (1 + a x)}{b} = 1$ $\frac{b (b x - 1)}{a} - \frac{a (1 + a x)}{b} = 1$
$\frac{x - 2 a}{b} - \frac{3}{2} = \frac{b - x}{2 a} + \frac{1}{2}$ $\frac{x - 2 a}{b} - \frac{3}{2} = \frac{b - x}{2 a} + \frac{1}{2}$

Critical Thinking/Discussion/Writing

Explain whether the equations in each pair are equivalent.
1. $x^{2} = x, x = 1$ $x^{2} = x, x = 1$
2. $x^{2} = 9, x = 3$ $x^{2} = 9, x = 3$
3. $x^{2} - 1 = x - 1, x = 0$ $x^{2} - 1 = x - 1, x = 0$
4. $\frac{x}{x - 2} = \frac{2}{x - 2}, x = 2$ $\frac{x}{x - 2} = \frac{2}{x - 2}, x = 2$
What is wrong with the following argument?

Let

$\begin{array}{rcll} x & = & 1 . \\ x^{2} & = & x & Multiply both sides by x \\ x^{2} - 1 & = & x - 1 & Subtract 1 from both sides . \\ \frac{x^{2} - 1}{x - 1} & = & \frac{x - 1}{x - 1} & Divide both sides by x - 1 . \\ \frac{(x + 1) (x - 1)}{x - 1} & = & 1 & Factor x^{2} - 1 . \\ x + 1 & = & 1 & Remove common factor . \\ 1 + 1 & = & 1 & Replace x with 1 in the \\ in the previous equation . \\ 2 & = & 1 & Add . \end{array}$ $\begin{array}{rcll} x & = & 1 . \\ x^{2} & = & x & Multiply both sides by x \\ x^{2} - 1 & = & x - 1 & Subtract 1 from both sides . \\ \frac{x^{2} - 1}{x - 1} & = & \frac{x - 1}{x - 1} & Divide both sides by x - 1 . \\ \frac{(x + 1) (x - 1)}{x - 1} & = & 1 & Factor x^{2} - 1 . \\ x + 1 & = & 1 & Remove common factor . \\ 1 + 1 & = & 1 & Replace x with 1 in the \\ in the previous equation . \\ 2 & = & 1 & Add . \end{array}$

Getting Ready for the Next Section

In Exercises 87–94, simplify each expression.

$(3 x + 7) - x + 2 (5 - x)$ $(3 x + 7) - x + 2 (5 - x)$
$- 4 (x - 3) - (1 - 2 x)$ $- 4 (x - 3) - (1 - 2 x)$
$- 2 x + 7 (x - 2) + 4$ $- 2 x + 7 (x - 2) + 4$
$6 (x - 1) + 2 (3 x - 5)$ $6 (x - 1) + 2 (3 x - 5)$
$\frac{2 x}{5} + \frac{x}{2}$ $\frac{2 x}{5} + \frac{x}{2}$
$\frac{4 x}{3} - \frac{9 x}{7}$ $\frac{4 x}{3} - \frac{9 x}{7}$
$\frac{3 x}{8} - \frac{2 x}{3}$ $\frac{3 x}{8} - \frac{2 x}{3}$
$\frac{5 x}{11} - \frac{x}{2}$ $\frac{5 x}{11} - \frac{x}{2}$

In Exercises 95–102, fill in the blank $(1 inch = 2.54 centimeters)$ $(1 inch = 2.54 centimeters)$ .

$24 minutes = \underline{} hours$ $24 minutes = \underline{} hours$
$75 minutes = \underline{} hours$ $75 minutes = \underline{} hours$
$0.25 hour = \underline{} minutes$ $0.25 hour = \underline{} minutes$
$2.5 hours = \underline{} minutes$ $2.5 hours = \underline{} minutes$
$5 yards = \underline{} feet$ $5 yards = \underline{} feet$
$1.4 yards = \underline{} feet$ $1.4 yards = \underline{} feet$
$3 inches = \underline{} centimeters$ $3 inches = \underline{} centimeters$
$10 inches = \underline{} centimeters$ $10 inches = \underline{} centimeters$

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Table of Contents for Section 1.1 Linear Equations in One Variable

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Table of Contents for
Section 1.1 Linear Equations in One Variable