Section3.3Isolating a Linear Variable
¶In this section, we will learn how to solve linear equations and inequalities with more than one variables.
Subsection3.3.1Solving for a Variable
The formula of calculating a rectangle's area is \(A=lw\text{,}\) where \(l\) 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 m^{2}, and it's length is given as 4 m, we could find its width this way:
\begin{align*} A\amp=lw\\ 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 \(A=lw\text{:}\)
\begin{align*} A\amp=lw\\ \divideunder{A}{l}\amp=\divideunder{lw}{l}\\ \frac{A}{l}\amp=w\\ w\amp=\frac{A}{l} \end{align*}Now if we want to find the width when \(l=4\) is given, we can simply replace \(l\) with \(4\) and simplify.
We solved for \(w\) in the formula \(A=lw\) once, and we could use the new formula \(w=\frac{A}{l}\) again and again saving us a lot of time down the road. Let's look at a few examples.
Remark3.3.2
Note that in solving for \(A\text{,}\) we divided each side of the equation by \(l\text{.}\) 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 \(A=lw\text{,}\) we saw that \(w\) was multiplied by \(l\text{,}\) and so we knew that in order to “undo” that operation, we would need to divide each side by \(l\text{.}\) We will see this process of “undoing” the operations throughout this section.
Example3.3.3
Solve for \(R\) in \(P=RC\text{.}\) (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*}Example3.3.4
Solve for \(x\) in \(y=mx+b\text{.}\) (This is a line's equation in slopeintercept 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}\\ yb\amp=m\attention{x} \end{align*}Now that we have \(mx\) on it's 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{yb}{m}\amp=\divideunder{m\attention{x}}{m}\\ \frac{yb}{m}\amp=\attention{x}\\ x\amp=\frac{yb}{m} \end{align*}Warning3.3.5
It's important to note in Example 3.3.4 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.
Example3.3.6
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*}Example3.3.7
Solve for \(y\) in \(2x+5y=10\text{.}\) (This is a linear equation in standard form.)
To solve for \(y\) in the equation \(2x+5y=10\text{,}\) we will first have to solve for \(5y\text{.}\) We'll do so 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=102x\\ \divideunder{5\attention{y}}{5}\amp=\divideunder{102x}{5}\\ y\amp=\frac{102x}{5} \end{align*}Remark3.3.8
As we will learn in later sections, the result in Example 3.3.7 can also be written as \(y=\frac{10}{5}\frac{2x}{5}\) which can then be written as \(y=2\frac{2}{5}x\text{.}\)
Example3.3.9
Solve for \(F\) in \(C=\frac{5}{9}(F32)\text{.}\) (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 \(F32\) 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 3.2 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 \(F32\text{,}\) 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*}Subsection3.3.2Exercises
Solving for a Variable
1

Solve this linear equation for \(y\text{.}\)
\(y+1=7\)

Solve this linear equation for \(x\text{.}\)
\(x + q = n\)
2

Solve this linear equation for \(y\text{.}\)
\(y+7=15\)

Solve this linear equation for \(t\text{.}\)
\(t + b = m\)
3

Solve this linear equation for \(r\text{.}\)
\(r5=3\)

Solve this linear equation for \(y\text{.}\)
\(ym=3\)
4

Solve this linear equation for \(r\text{.}\)
\(r7=1\)

Solve this linear equation for \(x\text{.}\)
\(xm=1\)
5

Solve this linear equation for \(t\text{.}\)
\(t+1=9\)

Solve this linear equation for \(r\text{.}\)
\(r+p=m\)
6

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

Solve this linear equation for \(y\text{.}\)
\(y+x=B\)
7

Solve this linear equation for \(t\text{.}\)
\(7t = 14\)

Solve this linear equation for \(x\text{.}\)
\(cx=b\)
8

Solve this linear equation for \(x\text{.}\)
\(7x = 42\)

Solve this linear equation for \(r\text{.}\)
\(pr=y\)
9

Solve this linear equation for \(x\text{.}\)
\(\frac{x}{7}=2\)

Solve this linear equation for \(y\text{.}\)
\(\frac{y}{t}=n\)
10

Solve this linear equation for \(y\text{.}\)
\(\frac{y}{7}=6\)

Solve this linear equation for \(x\text{.}\)
\(\frac{x}{C}=A\)
11

Solve this linear equation for \(y\text{.}\)
\(5y+8=48\)

Solve this linear equation for \(r\text{.}\)
\(tr+C=a\)
12

Solve this linear equation for \(r\text{.}\)
\(7r+9=44\)

Solve this linear equation for \(y\text{.}\)
\(Ay+q=C\)
13

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

Solve this linear equation for \(t\text{.}\)
\(rt = q\)
14

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

Solve this linear equation for \(r\text{.}\)
\(tr = b\)
15

Solve this linear equation for \(t\text{.}\)
\(t+y = m\)

Solve this linear equation for \(y\text{.}\)
\(t+y = m\)
16

Solve this linear equation for \(t\text{.}\)
\(t+x = p\)

Solve this linear equation for \(x\text{.}\)
\(t+x = p\)
17

Solve this linear equation for \(y\text{.}\)
\(Br+y=x\)

Solve this linear equation for \(B\text{.}\)
\(Br+y=x\)
18

Solve this linear equation for \(r\text{.}\)
\(tx+r=a\)

Solve this linear equation for \(t\text{.}\)
\(tx+r=a\)
19

Solve this linear equation for \(a\text{.}\)
\(y=pa+x\)

Solve this linear equation for \(p\text{.}\)
\(y=pa+x\)
20

Solve this linear equation for \(c\text{.}\)
\(t=Ac+m\)

Solve this linear equation for \(A\text{.}\)
\(t=Ac+m\)
21
Solve this linear equation for \(x\text{:}\)
\(\displaystyle{ y=mxb}\)
22
Solve this linear equation for \(x\text{:}\)
\(\displaystyle{ y=mx+b}\)
23

Solve this equation for \(b\text{:}\)
\(28=\frac{1}{2} b \cdot 8\)

Solve this equation for \(b\text{:}\)
\(A=\frac{1}{2} b \cdot h\)
24

Solve this equation for \(b\text{:}\)
\(20=\frac{1}{2} b \cdot 10\)

Solve this equation for \(b\text{:}\)
\(A=\frac{1}{2} b \cdot h\)
25
Solve this linear equation for \(r\text{:}\)
\(\displaystyle{ C=2 \pi r }\)
26
Solve this linear equation for \(h\text{:}\)
\(\displaystyle{ V= \pi r^{2} h }\)
27
Solve these linear equations for \(x\text{.}\)

\(\frac{x}{2}+5=9\)

\(\frac{x}{t}+5=n\)
28
Solve these linear equations for \(y\text{.}\)

\(\frac{y}{3}+6=9\)

\(\frac{y}{t}+6=r\)
29
Solve this linear equation for \(y\text{:}\)
\(\displaystyle{ \frac{y}{r}+t=b }\)
30
Solve this linear equation for \(r\text{:}\)
\(\displaystyle{ \frac{r}{x}+q=b }\)
31
Solve this linear equation for \(r\text{:}\)
\(\displaystyle{ \frac{r}{5}+y=C }\)
32
Solve this linear equation for \(t\text{:}\)
\(\displaystyle{ \frac{t}{5}+y=p }\)
33
Solve this linear equation for \(p\text{:}\)
\(\displaystyle{ r=a\frac{5p}{x} }\)
34
Solve this linear equation for \(q\text{:}\)
\(\displaystyle{ b=p\frac{6q}{C} }\)
35
Solve this linear equation for \(x\text{:}\)
\(\displaystyle{ Ax+By=C }\)
Note that the variables are upper case A, B, and C and lower case x and y.
36
Solve this linear equation for \(y\text{:}\)
\(\displaystyle{ Ax+By=C }\)
Note that the variables are upper case A, B, and C and lower case x and y.