ORCCA Isolating a Linear Variable

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Section 2.5 Isolating a Linear Variable

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Objectives: PCC Course Content and Outcome Guide

MTH 60 CCOG 2.6

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

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Figure 2.5.1. Alternative Video Lessons

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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.

permalinkBut 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{aligned} A & = ℓ w \\ 12 & = 4 w \\ \frac{12}{4} & = \frac{4 w}{4} \\ 3 & = w \\ w & = 3 \end{aligned}

$\begin{align*} A\amp=\ell w\\ 12\amp=4w\\ \divideunder{12}{4}\amp=\divideunder{4w}{4}\\ 3\amp=w\\ w\amp=3 \end{align*}$

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

\begin{aligned} A & = ℓ w \\ \frac{A}{ℓ} & = \frac{ℓ w}{ℓ} \\ \frac{A}{ℓ} & = w \\ w & = \frac{A}{ℓ} \end{aligned}

$\begin{align*} A\amp=\ell w\\ \divideunder{A}{\ell}\amp=\divideunder{\ell w}{\ell}\\ \frac{A}{\ell}\amp=w\\ w\amp=\frac{A}{\ell} \end{align*}$

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.

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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.

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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*}

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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*}

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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.

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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*}

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Checkpoint 2.5.7.

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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*}

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Reading Questions 2.5.2 Reading Questions

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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 $\frac{mq}{m}+b=\frac{T}{m}\text{?}$