## Section3.4Isolating a Linear Variable

###### ObjectivesPCC Course Content and Outcome Guide

In this section, we will learn how to solve linear equations and inequalities with more than one variable.

### Subsection3.4.1Solving for a Variable

The formula of calculating a rectangle's area is $A=\ell w\text{,}$ where $\ell$ stands for the rectangle's length, and $w$ stands for width. When a rectangle's length and width are given, we can easily calculate its area.

What if a rectangle's area and length are given, and we need to calculate its width?

If a rectangle's area is given as 12 m2, and its length is given as 4 m, we could find its width this way:

\begin{align*} A\amp=\ell w\\ 12\amp=4w\\ \divideunder{12}{4}\amp=\divideunder{4w}{4}\\ 3\amp=w\\ w\amp=3 \end{align*}

If we need to do this many times, we would love to have an easier way, without solving an equation each time. We will solve for $w$ in the formula $A=\ell w\text{:}$

\begin{align*} A\amp=\ell w\\ \divideunder{A}{\ell}\amp=\divideunder{\ell w}{\ell}\\ \frac{A}{\ell}\amp=w\\ w\amp=\frac{A}{\ell} \end{align*}

Now if we want to find the width when $\ell=4$ is given, we can simply replace $\ell$ with $4$ and simplify.

We solved for $w$ in the formula $A=\ell w$ once, and we could use the new formula $w=\frac{A}{\ell}$ again and again saving us a lot of time down the road. Let's look at a few examples.

###### Remark3.4.2

Note that in solving for $A\text{,}$ we divided each side of the equation by $\ell\text{.}$ The operations that we apply, and the order in which we do them, are determined by the operations in the original equation. In the original equation $A=\ell w\text{,}$ we saw that $w$ was multiplied by $\ell\text{,}$ and so we knew that in order to “undo” that operation, we would need to divide each side by $\ell\text{.}$ We will see this process of “un-doing” the operations throughout this section.

###### Example3.4.3

Solve for $R$ in $P=R-C\text{.}$ (This is the relationship between profit, revenue, and cost.)

Explanation

To solve for $R\text{,}$ we first want to note that $C$ is subtracted from $R\text{.}$ To “undo” this, we will need to add $C$ to each side of the equation:

###### Example3.4.4

Solve for $x$ in $y=mx+b\text{.}$ (This is a line's equation in slope-intercept form.)

Explanation

In the equation $y=mx+b\text{,}$ we see that $x$ is multiplied by $m$ and then $b$ is added to that. Our first step will be to isolate $mx\text{,}$ which we'll do by subtracting $b$ from each side of the equation:

\begin{align*} y\amp=m\attention{x}+b\\ y\subtractright{b}\amp=m\attention{x}+b\subtractright{b}\\ y-b\amp=m\attention{x} \end{align*}

Now that we have $mx$ on its own, we'll note that $x$ is multiplied by $m\text{.}$ To “undo” this, we'll need to divide each side of the equation by $m\text{:}$

\begin{align*} \divideunder{y-b}{m}\amp=\divideunder{m\attention{x}}{m}\\ \frac{y-b}{m}\amp=\attention{x}\\ x\amp=\frac{y-b}{m} \end{align*}
###### Warning3.4.5

It's important to note in Example 3.4.4 that each side was divided by $m\text{.}$ We can't simply divide $y$ by $m\text{,}$ as the equation would no longer be equivalent.

###### Example3.4.6

Solve for $b$ in $A=\frac{1}{2}bh\text{.}$ (This is the area formula for a triangle.)

Explanation

To solve for $b\text{,}$ we need to determine what operations need to be “undone.” The expression $\frac{1}{2}bh$ has multiplication between $\frac{1}{2}$ and $b$ and $h\text{.}$ As a first step, we will multiply each side of the equation by $2$ in order to eliminate the denominator of $2\text{:}$

\begin{align*} A\amp=\frac{1}{2}\attention{b}h\\ \multiplyleft{2}A\amp=\multiplyleft{2}\frac{1}{2}\attention{b}h\\ 2A\amp=\attention{b}h \end{align*}

As a last step, we will “undo” the multiplication between $b$ and $h$ by dividing each side by $h\text{:}$

\begin{align*} \divideunder{2A}{h}\amp=\divideunder{\attention{b}h}{h}\\ \frac{2A}{h}\amp=\attention{b}\\ b\amp=\frac{2A}{h} \end{align*}
###### Example3.4.7

Solve for $y$ in $2x+5y=10\text{.}$ (This is a linear equation in standard form.)

Explanation

To solve for $y\text{,}$ we will first have to solve for $5y$ by subtracting $2x$ from each side of the equation. After that, we'll be able to divide each side by $5$ to finish solving for $y\text{:}$

\begin{align*} 2x+5\attention{y}\amp=10\\ 2x+5\attention{y}\subtractright{2x}\amp=10\subtractright{2x}\\ 5\attention{y}\amp=10-2x\\ \divideunder{5\attention{y}}{5}\amp=\divideunder{10-2x}{5}\\ y\amp=\frac{10-2x}{5} \end{align*}
###### Remark3.4.8

As we will learn in later sections, the result in Example 3.4.7 can also be written as $y=\frac{10}{5}-\frac{2x}{5}$ which can then be written as $y=2-\frac{2}{5}x\text{.}$

###### Example3.4.9

Solve for $F$ in $C=\frac{5}{9}(F-32)\text{.}$ (This represents the relationship between temperature in degrees Celsius and degrees Fahrenheit.)

Explanation

To solve for $F\text{,}$ we first need to see that it is contained inside a set of parentheses. To get the expression $F-32$ on its own, we'll need to eliminate the $\frac{5}{9}$ outside those parentheses. One way we can “undo” this multiplication is to divide each side by $\frac{5}{9}\text{.}$ As we learned in Section 3.3 though, a better approach is to instead multiply each side by the reciprocal of $\frac{5}{9}\text{,}$ which is $\highlight{\frac{9}{5}}\text{:}$

\begin{align*} C\amp=\frac{5}{9}(\attention{F}-32)\\ \multiplyleft{\frac{9}{5}}C\amp=\multiplyleft{\frac{9}{5}}\frac{5}{9}(\attention{F}-32)\\ \frac{9}{5}C\amp=\attention{F}-32 \end{align*}

Now that we have $F-32\text{,}$ we simply need to add $32$ to each side to finish solving for $F\text{:}$

### Subsection3.4.2Exercises

###### 1

Solve the equation.

$\displaystyle{ {9q+2}={56} }$

###### 2

Solve the equation.

$\displaystyle{ {6y+1}={55} }$

###### 3

Solve the equation.

$\displaystyle{ {-9r-8}={37} }$

###### 4

Solve the equation.

$\displaystyle{ {-3a-5}={-17} }$

###### 5

Solve the equation.

$\displaystyle{ {-6b+3} = {-b-12} }$

###### 6

Solve the equation.

$\displaystyle{ {-2A+6} = {-A-4} }$

###### 7

Solve the equation.

$\displaystyle{ {56}={-8\!\left(B-10\right)} }$

###### 8

Solve the equation.

$\displaystyle{ {35}={-5\!\left(m-4\right)} }$

###### 9
1. Solve this linear equation for $t\text{.}$

$t+1=9$

2. Solve this linear equation for $x\text{.}$

$x + m = y$

###### 10
1. Solve this linear equation for $t\text{.}$

$t+1=5$

2. Solve this linear equation for $r\text{.}$

$r + n = y$

###### 11
1. Solve this linear equation for $x\text{.}$

$x-5=-3$

2. Solve this linear equation for $y\text{.}$

$y-B=-3$

###### 12
1. Solve this linear equation for $x\text{.}$

$x-5=-3$

2. Solve this linear equation for $y\text{.}$

$y-c=-3$

###### 13
1. Solve this linear equation for $y\text{.}$

$-y+1=-5$

2. Solve this linear equation for $r\text{.}$

$-r+c=q$

###### 14
1. Solve this linear equation for $y\text{.}$

$-y+7=3$

2. Solve this linear equation for $t\text{.}$

$-t+C=b$

###### 15
1. Solve this linear equation for $r\text{.}$

$7r = 56$

2. Solve this linear equation for $t\text{.}$

$ct=x$

###### 16
1. Solve this linear equation for $r\text{.}$

$3r = 12$

2. Solve this linear equation for $x\text{.}$

$yx=n$

###### 17
1. Solve this linear equation for $r\text{.}$

$\frac{r}{7}=10$

2. Solve this linear equation for $y\text{.}$

$\frac{y}{p}=x$

###### 18
1. Solve this linear equation for $t\text{.}$

$\frac{t}{3}=2$

2. Solve this linear equation for $r\text{.}$

$\frac{r}{B}=x$

###### 19
1. Solve this linear equation for $t\text{.}$

$7t+5=12$

2. Solve this linear equation for $r\text{.}$

$qr+a=x$

###### 20
1. Solve this linear equation for $x\text{.}$

$6x+4=64$

2. Solve this linear equation for $y\text{.}$

$qy+r=n$

###### 21
1. Solve this linear equation for $x\text{.}$

$xt = m$

2. Solve this linear equation for $t\text{.}$

$xt = m$

###### 22
1. Solve this linear equation for $y\text{.}$

$yr = c$

2. Solve this linear equation for $r\text{.}$

$yr = c$

###### 23
1. Solve this linear equation for $y\text{.}$

$y+x = B$

2. Solve this linear equation for $x\text{.}$

$y+x = B$

###### 24
1. Solve this linear equation for $r\text{.}$

$r+t = x$

2. Solve this linear equation for $t\text{.}$

$r+t = x$

###### 25
1. Solve this linear equation for $B\text{.}$

$cy+B=C$

2. Solve this linear equation for $c\text{.}$

$cy+B=C$

###### 26
1. Solve this linear equation for $m\text{.}$

$rt+m=b$

2. Solve this linear equation for $r\text{.}$

$rt+m=b$

###### 27
1. Solve this linear equation for $n\text{.}$

$x=tn+p$

2. Solve this linear equation for $t\text{.}$

$x=tn+p$

###### 28
1. Solve this linear equation for $q\text{.}$

$r=cq+n$

2. Solve this linear equation for $c\text{.}$

$r=cq+n$

###### 29

Solve this linear equation for $x\text{.}$

\begin{equation*} y=mx-b \end{equation*}
###### 30

Solve this linear equation for $x\text{.}$

\begin{equation*} y=-mx+b \end{equation*}
###### 31
1. Solve this equation for $b\text{:}$

$12=\frac{1}{2} b \cdot 4$

2. Solve this equation for $b\text{:}$

$A=\frac{1}{2} b \cdot h$

###### 32
1. Solve this equation for $b\text{:}$

$9=\frac{1}{2} b \cdot 6$

2. Solve this equation for $b\text{:}$

$A=\frac{1}{2} b \cdot h$

###### 33

Solve this linear equation for $r\text{.}$

\begin{equation*} C=2 \pi r \end{equation*}
###### 34

Solve this linear equation for $h\text{.}$

\begin{equation*} V= \pi r^{2} h \end{equation*}
###### 35

Solve these linear equations for $r\text{.}$

1. $\frac{r}{5}+9=12$

2. $\frac{r}{x}+9=B$

###### 36

Solve these linear equations for $t\text{.}$

1. $\frac{t}{5}+7=8$

2. $\frac{t}{r}+7=x$

###### 37

Solve this linear equation for $t\text{.}$

\begin{equation*} \frac{t}{y}+c=a \end{equation*}
###### 38

Solve this linear equation for $x\text{.}$

\begin{equation*} \frac{x}{t}+p=m \end{equation*}
###### 39

Solve this linear equation for $x\text{.}$

\begin{equation*} \frac{x}{9}+r=a \end{equation*}
###### 40

Solve this linear equation for $y\text{.}$

\begin{equation*} \frac{y}{8}+r=A \end{equation*}
###### 41

Solve this linear equation for $b\text{.}$

\begin{equation*} t=y-\frac{8b}{q} \end{equation*}
###### 42

Solve this linear equation for $A\text{.}$

\begin{equation*} C=q-\frac{2A}{B} \end{equation*}
###### 43

Solve this linear equation for $x\text{.}$

\begin{equation*} Ax+By=C \end{equation*}
###### 44

Solve this linear equation for $y\text{.}$

\begin{equation*} Ax+By=C \end{equation*}
###### 45

Solve the linear equation for $y\text{.}$

\begin{equation*} {25x+5y}={-75} \end{equation*}
###### 46

Solve the linear equation for $y\text{.}$

\begin{equation*} {30x-5y}={-65} \end{equation*}
###### 47

Solve the linear equation for $y\text{.}$

\begin{equation*} {4x+2y}={6} \end{equation*}
###### 48

Solve the linear equation for $y\text{.}$

\begin{equation*} {18x-2y}={16} \end{equation*}
###### 49

Solve the linear equation for $y\text{.}$

\begin{equation*} {4x-y}={14} \end{equation*}
###### 50

Solve the linear equation for $y\text{.}$

\begin{equation*} {2x-y}={-12} \end{equation*}
###### 51

Solve the linear equation for $y\text{.}$

\begin{equation*} {-4x-6y}={-24} \end{equation*}
###### 52

Solve the linear equation for $y\text{.}$

\begin{equation*} {-7x-6y}={-18} \end{equation*}
###### 53

Solve the linear equation for $y\text{.}$

\begin{equation*} {2x+7y}={2} \end{equation*}
###### 54

Solve the linear equation for $y\text{.}$

\begin{equation*} {6x-8y}={2} \end{equation*}
###### 55

Solve the linear equation for $y\text{.}$

\begin{equation*} {-87x-87y}={38} \end{equation*}
###### 56

Solve the linear equation for $y\text{.}$

\begin{equation*} {24y-46x}={25} \end{equation*}