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Section3.3Isolating a Linear Variable

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

Figure3.3.1Alternative Video Lessons

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*}

Subsection3.3.2Exercises

Solving for a Variable

1

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

    \(y+1=7\)

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

    \(x + q = n\)

2

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

    \(y+7=15\)

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

    \(t + b = m\)

3

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

    \(r-5=3\)

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

    \(y-m=3\)

4

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

    \(r-7=1\)

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

    \(x-m=1\)

5

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

    \(-t+1=-9\)

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

    \(-r+p=m\)

6

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

    \(-t+7=1\)

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

    \(-y+x=B\)

7

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

    \(7t = 14\)

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

    \(cx=b\)

8

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

    \(7x = 42\)

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

    \(pr=y\)

9

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

    \(\frac{x}{7}=2\)

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

    \(\frac{y}{t}=n\)

10

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

    \(\frac{y}{7}=6\)

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

    \(\frac{x}{C}=A\)

11

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

    \(5y+8=48\)

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

    \(tr+C=a\)

12

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

    \(7r+9=44\)

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

    \(Ay+q=C\)

13

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

    \(rt = q\)

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

    \(rt = q\)

14

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

    \(tr = b\)

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

    \(tr = b\)

15

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

    \(t+y = m\)

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

    \(t+y = m\)

16

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

    \(t+x = p\)

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

    \(t+x = p\)

17

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

    \(Br+y=x\)

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

    \(Br+y=x\)

18

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

    \(tx+r=a\)

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

    \(tx+r=a\)

19

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

    \(y=pa+x\)

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

    \(y=pa+x\)

20

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

    \(t=Ac+m\)

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

    \(t=Ac+m\)

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

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

    \(28=\frac{1}{2} b \cdot 8\)

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

    \(A=\frac{1}{2} b \cdot h\)

24

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

    \(20=\frac{1}{2} b \cdot 10\)

  2. 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 \(x\text{.}\)

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

  2. \(\frac{x}{t}+5=n\)

28

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

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

  2. \(\frac{y}{t}+6=r\)

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.