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 m2, 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 “un-doing” the operations throughout this section.
Example3.3.3
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*}
Example3.3.4
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 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{y-b}{m}\amp=\divideunder{m\attention{x}}{m}\\
\frac{y-b}{m}\amp=\attention{x}\\
x\amp=\frac{y-b}{m}
\end{align*}
Warning3.3.5
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=10-2x\\
\divideunder{5\attention{y}}{5}\amp=\divideunder{10-2x}{5}\\
y\amp=\frac{10-2x}{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 \(C=\frac{5}{9}(F-32)\text{.}\) (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 3.2 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 \(F-32\text{,}\) 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*}
SubsectionExercises
Solving for a Variable
1
-
Solve this linear equation for \(t\text{.}\)
\(t+7=13\)
-
Solve this linear equation for \(y\text{.}\)
\(y + x = B\)
2
-
Solve this linear equation for \(t\text{.}\)
\(t+7=9\)
-
Solve this linear equation for \(x\text{.}\)
\(x + c = b\)
3
-
Solve this linear equation for \(x\text{.}\)
\(x-1=7\)
-
Solve this linear equation for \(r\text{.}\)
\(r-p=7\)
4
-
Solve this linear equation for \(x\text{.}\)
\(x-9=-1\)
-
Solve this linear equation for \(y\text{.}\)
\(y-t=-1\)
5
-
Solve this linear equation for \(y\text{.}\)
\(-y+7=1\)
-
Solve this linear equation for \(x\text{.}\)
\(-x+C=A\)
6
-
Solve this linear equation for \(y\text{.}\)
\(-y+1=-3\)
-
Solve this linear equation for \(r\text{.}\)
\(-r+t=C\)
7
-
Solve this linear equation for \(r\text{.}\)
\(7r = 42\)
-
Solve this linear equation for \(y\text{.}\)
\(Ay=q\)
8
-
Solve this linear equation for \(r\text{.}\)
\(7r = 14\)
-
Solve this linear equation for \(t\text{.}\)
\(qt=C\)
9
-
Solve this linear equation for \(t\text{.}\)
\(\frac{t}{7}=8\)
-
Solve this linear equation for \(r\text{.}\)
\(\frac{r}{b}=B\)
10
-
Solve this linear equation for \(t\text{.}\)
\(\frac{t}{3}=8\)
-
Solve this linear equation for \(y\text{.}\)
\(\frac{y}{m}=B\)
11
-
Solve this linear equation for \(t\text{.}\)
\(9t+4=49\)
-
Solve this linear equation for \(x\text{.}\)
\(px+B=A\)
12
-
Solve this linear equation for \(x\text{.}\)
\(7x+5=40\)
-
Solve this linear equation for \(r\text{.}\)
\(Br+p=n\)
13
-
Solve this linear equation for \(x\text{.}\)
\(xy = a\)
-
Solve this linear equation for \(y\text{.}\)
\(xy = a\)
14
-
Solve this linear equation for \(y\text{.}\)
\(yt = c\)
-
Solve this linear equation for \(t\text{.}\)
\(yt = c\)
15
-
Solve this linear equation for \(y\text{.}\)
\(y+r = p\)
-
Solve this linear equation for \(r\text{.}\)
\(y+r = p\)
16
-
Solve this linear equation for \(r\text{.}\)
\(r+x = a\)
-
Solve this linear equation for \(x\text{.}\)
\(r+x = a\)
17
-
Solve this linear equation for \(C\text{.}\)
\(pr+C=A\)
-
Solve this linear equation for \(p\text{.}\)
\(pr+C=A\)
18
-
Solve this linear equation for \(m\text{.}\)
\(Ax+m=p\)
-
Solve this linear equation for \(A\text{.}\)
\(Ax+m=p\)
19
-
Solve this linear equation for \(p\text{.}\)
\(y=rp+A\)
-
Solve this linear equation for \(r\text{.}\)
\(y=rp+A\)
20
-
Solve this linear equation for \(q\text{.}\)
\(t=nq+b\)
-
Solve this linear equation for \(n\text{.}\)
\(t=nq+b\)
21
Solve this linear equation for \(x\text{:}\)
\(\displaystyle{ y=mx-b}\)
22
Solve this linear equation for \(x\text{:}\)
\(\displaystyle{ y=-mx+b}\)
23
-
Solve this equation for \(b\text{:}\)
\(18=\frac{1}{2} b \cdot 4\)
-
Solve this equation for \(b\text{:}\)
\(A=\frac{1}{2} b \cdot h\)
24
-
Solve this equation for \(b\text{:}\)
\(18=\frac{1}{2} b \cdot 6\)
-
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 \(t\text{.}\)
\(\frac{t}{3}+5=8\)
\(\frac{t}{y}+5=p\)
28
Solve these linear equations for \(t\text{.}\)
\(\frac{t}{3}+2=3\)
\(\frac{t}{x}+2=a\)
29
Solve this linear equation for \(t\text{:}\)
\(\displaystyle{ \frac{t}{r}+C=p }\)
30
Solve this linear equation for \(x\text{:}\)
\(\displaystyle{ \frac{x}{y}+B=a }\)
31
Solve this linear equation for \(x\text{:}\)
\(\displaystyle{ \frac{x}{5}+r=a }\)
32
Solve this linear equation for \(y\text{:}\)
\(\displaystyle{ \frac{y}{2}+r=q }\)
33
Solve this linear equation for \(b\text{:}\)
\(\displaystyle{ A=m-\frac{8b}{a} }\)
34
Solve this linear equation for \(A\text{:}\)
\(\displaystyle{ c=n-\frac{4A}{q} }\)
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.