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Section 2.5 Isolating a Linear Variable

In this section, we solve for a variable in a linear equation, even when there is more than one variable.

Figure 2.5.1. Alternative Video Lessons

Subsection 2.5.1 Solving 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. So when a rectangle's length and width are given, we can easily calculate its area.

But what if we know a rectangle's area and length, and we need to calculate its width?

If a rectangle's area is 12 m2, and its length is 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*}

We can go through the same motions without using the specific numbers for \(A\) and \(\ell\text{:}\)

\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 \(A=12\) and \(\ell=4\text{,}\) we have a formula: \(w=\frac{A}{\ell}\text{.}\) But the formula could also quickly tell us the width when \(A=100\) and \(\ell=20\text{,}\) or when when \(A=23.47\) and \(\ell=2.71\text{,}\) or any other variation. This formula, \(w=\frac{A}{\ell}\text{,}\) is a handy version of the original equation in situations where \(w\) is the unknown.

Remark 2.5.2.

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 continue to use this process of “un-doing” operations throughout this section.

Example 2.5.3.

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 add \(C\) to each side of the equation:

\begin{align*} P\amp=\attention{R}-C\\ P\addright{C}\amp=\attention{R}-C\addright{C}\\ P+C\amp=\attention{R}\\ R\amp=P+C \end{align*}
Example 2.5.4.

Solve for \(x\) in \(y=mx+b\text{.}\) (This is a line's equation in slope-intercept form, studied more in Section 3.5.)

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 \(mx\) by itself, we note that \(x\) is multiplied by \(m\text{.}\) To undo this, we 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*}
Warning 2.5.5.

It's important to note in Example 2.5.4 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.

Example 2.5.6.

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 \(\frac{1}{2}\text{:}\)

\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*}

Next we 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*}
Checkpoint 2.5.7.
Example 2.5.8.

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 \(F\text{,}\) note that it is contained inside parentheses. To isolate the expression \(F-32\text{,}\) we want to eliminate the \(\frac{5}{9}\) outside those parentheses. One way we can undo this multiplication is to divide each side by \(\frac{5}{9}\text{.}\) A better technique is to multiply each side by the reciprocal of \(\frac{5}{9}\text{,}\) which is \(\frac{9}{5}\text{:}\)

\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 \(F\text{:}\)

\begin{align*} \frac{9}{5}C\addright{32}\amp=\attention{F}-32\addright{32}\\ \frac{9}{5}C+32\amp=\attention{F}\\ F\amp=\frac{9}{5}C+32 \end{align*}

Reading Questions 2.5.2 Reading Questions

1.

Suppose you want to solve the equation \(mq+b=T\) for \(q\text{.}\) What would be wrong with dividing on each side by \(m\) to get \(\frac{mq}{m}+b=\frac{T}{m}\text{?}\)

2.

How do you undo dividing by \(R\text{?}\)

Exercises 2.5.3 Exercises

Review and Warmup
1.

Solve the equation.

\(\displaystyle{ {8q+4}={52} }\)

2.

Solve the equation.

\(\displaystyle{ {5y+3}={53} }\)

3.

Solve the equation.

\(\displaystyle{ {-10r-1}={29} }\)

4.

Solve the equation.

\(\displaystyle{ {-4a-9}={-29} }\)

5.

Solve the equation.

\(\displaystyle{ {-5b+3} = {-b-21} }\)

6.

Solve the equation.

\(\displaystyle{ {-10A+7} = {-A-20} }\)

7.

Solve the equation.

\(\displaystyle{ {-63}={-7\!\left(B-1\right)} }\)

8.

Solve the equation.

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

Solving for a Variable
9.

  1. Solve \(t+5=13\) for \(t\text{.}\)

  2. Solve \(y + r = p\) for \(y\text{.}\)

10.

  1. Solve \(t+1=7\) for \(t\text{.}\)

  2. Solve \(r + b = c\) for \(r\text{.}\)

11.

  1. Solve \(x-9=-7\) for \(x\text{.}\)

  2. Solve \(y-C=-7\) for \(y\text{.}\)

12.

  1. Solve \(x-5=-3\) for \(x\text{.}\)

  2. Solve \(t-y=-3\) for \(t\text{.}\)

13.

  1. Solve \(-y+1=-9\) for \(y\text{.}\)

  2. Solve \(-r+A=a\) for \(r\text{.}\)

14.

  1. Solve \(-y+3=-7\) for \(y\text{.}\)

  2. Solve \(-x+q=r\) for \(x\text{.}\)

15.

  1. Solve \(3r = 30\) for \(r\text{.}\)

  2. Solve \(at=m\) for \(t\text{.}\)

16.

  1. Solve \(5r = 10\) for \(r\text{.}\)

  2. Solve \(ct=y\) for \(t\text{.}\)

17.

  1. Solve \(\frac{r}{3}=10\) for \(r\text{.}\)

  2. Solve \(\frac{x}{y}=n\) for \(x\text{.}\)

18.

  1. Solve \(\frac{t}{5}=4\) for \(t\text{.}\)

  2. Solve \(\frac{r}{q}=y\) for \(r\text{.}\)

19.

  1. Solve \(4t+3=35\) for \(t\text{.}\)

  2. Solve \(xy+C=n\) for \(y\text{.}\)

20.

  1. Solve \(6x+6=24\) for \(x\text{.}\)

  2. Solve \(by+a=r\) for \(y\text{.}\)

21.

  1. Solve \(xt = n\) for \(x\text{.}\)

  2. Solve \(xt = n\) for \(t\text{.}\)

22.

  1. Solve \(yt = x\) for \(y\text{.}\)

  2. Solve \(yt = x\) for \(t\text{.}\)

23.

  1. Solve \(y+x = C\) for \(y\text{.}\)

  2. Solve \(y+x = C\) for \(x\text{.}\)

24.

  1. Solve \(r+x = t\) for \(r\text{.}\)

  2. Solve \(r+x = t\) for \(x\text{.}\)

25.

  1. Solve \(br+B=q\) for \(B\text{.}\)

  2. Solve \(br+B=q\) for \(b\text{.}\)

26.

  1. Solve \(xt+m=C\) for \(m\text{.}\)

  2. Solve \(xt+m=C\) for \(x\text{.}\)

27.

  1. Solve \(y=Cn+c\) for \(n\text{.}\)

  2. Solve \(y=Cn+c\) for \(C\text{.}\)

28.

  1. Solve \(t=aq+r\) for \(q\text{.}\)

  2. Solve \(t=aq+r\) for \(a\text{.}\)

29.

  1. Solve \(10=\frac{1}{2} b \cdot 2\) for \(b\text{.}\)

  2. Solve \(A=\frac{1}{2} b \cdot h\) for \(b\text{.}\)

30.

  1. Solve \(8=\frac{1}{2} b \cdot 2\) for \(b\text{.}\)

  2. Solve \(A=\frac{1}{2} b \cdot h\) for \(b\text{.}\)

31.

Solve these linear equations for \(y\text{.}\)

  1. \(\frac{y}{5}+10=12\)

  2. \(\frac{y}{x}+10=a\)

32.

Solve these linear equations for \(y\text{.}\)

  1. \(\frac{y}{5}+1=3\)

  2. \(\frac{y}{t}+1=c\)

33.

Solve this linear equation for \(x\text{.}\)

\begin{equation*} y=mx-b \end{equation*}
34.

Solve this linear equation for \(x\text{.}\)

\begin{equation*} y=-mx+b \end{equation*}
35.

Solve this linear equation for \(r\text{.}\)

\begin{equation*} C=2 \pi r \end{equation*}
36.

Solve this linear equation for \(h\text{.}\)

\begin{equation*} V= \pi r^{2} h \end{equation*}
37.

Solve this linear equation for \(t\text{.}\)

\begin{equation*} \frac{t}{x}+A=C \end{equation*}
38.

Solve this linear equation for \(x\text{.}\)

\begin{equation*} \frac{x}{t}+q=b \end{equation*}
39.

Solve this linear equation for \(x\text{.}\)

\begin{equation*} \frac{x}{2}+r=a \end{equation*}
40.

Solve this linear equation for \(y\text{.}\)

\begin{equation*} \frac{y}{9}+x=m \end{equation*}
41.

Solve this linear equation for \(b\text{.}\)

\begin{equation*} A=r-\frac{9b}{n} \end{equation*}
42.

Solve this linear equation for \(A\text{.}\)

\begin{equation*} t=B-\frac{9A}{c} \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 \(y\text{.}\)

\({-35x-5y}={10}\)

46.

Solve the linear equation for \(y\text{.}\)

\({20x-5y}={65}\)

47.

Solve the linear equation for \(y\text{.}\)

\({8x+2y}={30}\)

48.

Solve the linear equation for \(y\text{.}\)

\({14x-2y}={-10}\)

49.

Solve the linear equation for \(y\text{.}\)

\({4x-y}={18}\)

50.

Solve the linear equation for \(y\text{.}\)

\({2x-y}={-8}\)

51.

Solve the linear equation for \(y\text{.}\)

\({4x-6y}={-36}\)

52.

Solve the linear equation for \(y\text{.}\)

\({8x-6y}={-36}\)

53.

Solve the linear equation for \(y\text{.}\)

\({7y-3x}={5}\)

54.

Solve the linear equation for \(y\text{.}\)

\({6x-8y}={5}\)

55.

Solve the linear equation for \(y\text{.}\)

\({21x+87y}={110}\)

56.

Solve the linear equation for \(y\text{.}\)

\({24y-51x}={97}\)

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