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Section 3.4 Isolating a Linear Variable

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

Subsection 3.4.1 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. 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.

Remark 3.4.1.

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.

Example 3.4.2.

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

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:

\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 3.4.3.

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

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*}
Warning 3.4.4.

It's important to note in Example 3.4.3 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 3.4.5.

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

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*}
Example 3.4.6.

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

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*}
Remark 3.4.7.

As we will learn in later sections, the result in Example 3.4.6 can also be written as \(y=\frac{10}{5}-\frac{2x}{5}\text{,}\) which can then be written as \(y=2-\frac{2}{5}x\text{.}\)

Example 3.4.8.

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

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 by dividing 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 \(\frac{9}{5}\text{:}\)

\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*}

Exercises 3.4.2 Exercises

Review and Warmup
1.

Solve the equation.

\(\displaystyle{ {7r+5}={33} }\)

2.

Solve the equation.

\(\displaystyle{ {4a+4}={36} }\)

3.

Solve the equation.

\(\displaystyle{ {-2b-3}={13} }\)

4.

Solve the equation.

\(\displaystyle{ {-5A-10}={-15} }\)

5.

Solve the equation.

\(\displaystyle{ {-4B+10} = {-B-11} }\)

6.

Solve the equation.

\(\displaystyle{ {-9m+5} = {-m-27} }\)

7.

Solve the equation.

\(\displaystyle{ {96}={-6\!\left(n-8\right)} }\)

8.

Solve the equation.

\(\displaystyle{ {-12}={-3\!\left(q-3\right)} }\)

9.

Solve the equation.

\(\displaystyle{ {54}={-9\!\left(x-7\right)} }\)

10.

Solve the equation.

\(\displaystyle{ {\frac{r}{6}+3}={5} }\)

11.

Solve the equation.

\(\displaystyle{ {\frac{a}{3}+1}={5} }\)

12.

Solve the equation.

\(\displaystyle{ {9-\frac{b}{9}} = {4} }\)

13.

Solve the equation.

\(\displaystyle{ {5-\frac{A}{4}} = {1} }\)

14.

Solve the equation.

\(\displaystyle{ {-6} = {2-\frac{2B}{7}} }\)

15.

Solve the equation.

\(\displaystyle{ {-32} = {8-\frac{10m}{3}} }\)

Solving for a Variable
16.

  1. Solve \(t+5=15\) for \(t\text{.}\)

  2. Solve \(y + A = B\) for \(y\text{.}\)

17.

  1. Solve \(t+5=11\) for \(t\text{.}\)

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

18.

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

  2. Solve \(r-b=5\) for \(r\text{.}\)

19.

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

  2. Solve \(y-n=-1\) for \(y\text{.}\)

20.

  1. Solve \(-y+5=-5\) for \(y\text{.}\)

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

21.

  1. Solve \(-y+5=-1\) for \(y\text{.}\)

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

22.

  1. Solve \(5r = 50\) for \(r\text{.}\)

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

23.

  1. Solve \(5r = 30\) for \(r\text{.}\)

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

24.

  1. Solve \(\frac{r}{9}=10\) for \(r\text{.}\)

  2. Solve \(\frac{t}{b}=c\) for \(t\text{.}\)

25.

  1. Solve \(\frac{t}{5}=2\) for \(t\text{.}\)

  2. Solve \(\frac{y}{a}=c\) for \(y\text{.}\)

26.

  1. Solve \(2t+9=23\) for \(t\text{.}\)

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

27.

  1. Solve \(5x+2=12\) for \(x\text{.}\)

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

28.

  1. Solve \(xr = a\) for \(x\text{.}\)

  2. Solve \(xr = a\) for \(r\text{.}\)

29.

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

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

30.

  1. Solve \(y+r = b\) for \(y\text{.}\)

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

31.

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

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

32.

  1. Solve \(nx+B=r\) for \(B\text{.}\)

  2. Solve \(nx+B=r\) for \(n\text{.}\)

33.

  1. Solve \(Ar+m=p\) for \(m\text{.}\)

  2. Solve \(Ar+m=p\) for \(A\text{.}\)

34.

  1. Solve \(x=rn+B\) for \(n\text{.}\)

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

35.

  1. Solve \(y=nq+b\) for \(q\text{.}\)

  2. Solve \(y=nq+b\) for \(n\text{.}\)

36.

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

\begin{equation*} y=mx-b \end{equation*}
37.

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

\begin{equation*} y=-mx+b \end{equation*}
38.

  1. Solve \(18=\frac{1}{2} b \cdot 4\) for \(b\text{.}\)

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

39.

  1. Solve \(12=\frac{1}{2} b \cdot 4\) for \(b\text{.}\)

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

40.

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

\begin{equation*} C=2 \pi r \end{equation*}
41.

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

\begin{equation*} V= \pi r^{2} h \end{equation*}
42.

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

  1. \(\frac{r}{2}+5=10\)

  2. \(\frac{r}{y}+5=b\)

43.

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

  1. \(\frac{t}{2}+2=4\)

  2. \(\frac{t}{x}+2=n\)

44.

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

\begin{equation*} \frac{t}{r}+p=C \end{equation*}
45.

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

\begin{equation*} \frac{x}{y}+B=r \end{equation*}
46.

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

\begin{equation*} \frac{x}{3}+t=a \end{equation*}
47.

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

\begin{equation*} \frac{y}{2}+r=c \end{equation*}
48.

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

\begin{equation*} m=p-\frac{3b}{a} \end{equation*}
49.

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

\begin{equation*} y=a-\frac{5c}{q} \end{equation*}
50.

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

\begin{equation*} Ax+By=C \end{equation*}
51.

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

\begin{equation*} Ax+By=C \end{equation*}
52.

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

\begin{equation*} {5x+5y}={30} \end{equation*}
53.

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

\begin{equation*} {45x+5y}={-25} \end{equation*}
54.

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

\begin{equation*} {4x-2y}={32} \end{equation*}
55.

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

\begin{equation*} {12x+2y}={24} \end{equation*}
56.

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

\begin{equation*} {4x-y}={11} \end{equation*}
57.

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

\begin{equation*} {2x-y}={13} \end{equation*}
58.

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

\begin{equation*} {2x+5y}={-95} \end{equation*}
59.

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

\begin{equation*} {5x-6y}={-114} \end{equation*}
60.

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

\begin{equation*} {-9x-7y}={19} \end{equation*}
61.

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

\begin{equation*} {8y-4x}={18} \end{equation*}
62.

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

\begin{equation*} {70x-86y}={452} \end{equation*}
63.

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

\begin{equation*} {29x-23y}={439} \end{equation*}