11.4 Solving equations of the form ax=b and x/a=b

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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses solving equations of the form $ax=b$ and $\frac{x}{a}=b$ . By the end of the module students should be familiar with the multiplication/division property of equality, be able to solve equations of the form $\text{ax}=b$ and $\frac{x}{a}=b$ and be able to use combined techniques to solve equations.

Section overview

• Multiplication/ Division Property of Equality
• Combining Techniques in Equations Solving

Multiplication/ division property of equality

Recall that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side. From this, we can suggest the multiplication/division property of equality.

Multiplication/division property of equality

Given any equation,

1. We can obtain an equivalent equation by multiplying both sides of the equa­tion by the same nonzero number, that is, if $c\ne 0$ , then $a=b$ is equivalent to
$a\cdot c=b\cdot c$
2. We can obtain an equivalent equation by dividing both sides of the equation by the same nonzero number , that is, if $c\ne 0$ , then $a=b$ is equivalent to
$\frac{a}{c}=\frac{b}{c}$

The multiplication/division property of equality can be used to undo an association with a number that multiplies or divides the variable.

Sample set a

Use the multiplication / division property of equality to solve each equation.

$6y=\text{54}$
6 is associated with y by multiplication. Undo the association by dividing both sides by 6

$\begin{array}{c}\frac{6y}{6}=\frac{\text{54}}{6}\\ \frac{\overline{)6}y}{\overline{)6}}=\frac{\stackrel{9}{\overline{)54}}}{\overline{)6}}\\ y=9\end{array}$

Check: When $y=9$

$6y=\text{54}$

becomes
,
a true statement.

The solution to $6y=\text{54}$ is $y=9$ .

$\frac{x}{-2}=\text{27}$ .
-2 is associated with $x$ by division. Undo the association by multiplying both sides by -2.

$\begin{array}{}\left(-2\right)\frac{x}{-2}=\left(-2\right)\text{27}\\ \end{array}$

$\begin{array}{}\left(\overline{)-2}\right)\frac{x}{\overline{)-2}}=\left(-2\right)\text{27}\\ \end{array}$

$x=-\text{54}$

Check: When $x=-\text{54}$ ,

$\frac{x}{-2}=\text{27}$

becomes

a true statement.

The solution to $\frac{x}{-2}=\text{27}$ is $x=-\text{54}$

$\frac{3a}{7}=6$ .
We will examine two methods for solving equations such as this one.

Method 1: Use of dividing out common factors.

$\frac{3a}{7}=6$
7 is associated with $a$ by division. Undo the association by multiplying both sides by 7.

$7\cdot \frac{3a}{7}=7\cdot 6$
Divide out the 7’s.

$\overline{)7}\cdot \frac{3a}{\overline{)7}}=\text{42}$

$3a=\text{42}$
3 is associated with $a$ by multiplication. Undo the association by dviding both sides by 3.

$\frac{3a}{3}=\frac{\text{42}}{3}$

$\frac{\overline{)3}a}{\overline{)3}}=\text{14}$

$a=\text{14}$

Check: When $a=\text{14}$ ,

$\frac{3a}{7}=6$

becomes
,
a true statement.

The solution to $\frac{3a}{7}=6$ is $a=\text{14}$ .

Method 2: Use of reciprocals

Recall that if the product of two numbers is 1, the numbers are reciprocals . Thus $\frac{3}{7}$ and $\frac{7}{3}$ are reciprocals.

$\frac{3a}{7}=6$
Multiply both sides of the equation by $\frac{7}{3}$ , the reciprocal of $\frac{3}{7}$ .

$\frac{7}{3}\cdot \frac{3a}{7}=\frac{7}{3}\cdot 6$

$\frac{\stackrel{1}{\overline{)7}}}{\underset{1}{\overline{)3}}}\cdot \frac{\stackrel{1}{\overline{)3}a}}{\underset{1}{\overline{)7}}}=\frac{7}{\underset{1}{\overline{)3}}}\cdot \frac{\stackrel{2}{\overline{)6}}}{1}$

$\begin{array}{c}1\cdot a=\text{14}\hfill \\ a=\text{14}\hfill \end{array}$

Notice that we get the same solution using either method.

$-8x=\text{24}$
-8 is associated with $x$ by multiplication. Undo the association by dividing both sides by -8.

$\begin{array}{c}\frac{-8x}{-8}=\frac{\text{24}}{-8}\hfill \\ \hfill \end{array}$

$\frac{-8x}{-8}=\frac{\text{24}}{-8}$

$x=-3$

Check: When $x=-3$ ,

$-8x=\text{24}$

becomes
,
a true statement.

$-x=7\text{.}$
Since $–x$ is actually $-1\cdot x$ and $\left(-1\right)\left(-1\right)=1$ , we can isolate $x$ by multiplying both sides of the equation by $-1$ .

$\begin{array}{c}\left(-1\right)\left(-x\right)=-1\cdot 7\hfill \\ x=-7\hfill \end{array}$

Check: When $x=7$ ,

$-x=7$

becomes

The solution to $-x=7$ is $x=-7$ .

Practice set a

Use the multiplication/division property of equality to solve each equation. Be sure to check each solution.

$7x=\text{21}$

$x=3$

$-5x=\text{65}$

$x=-\text{13}$

$\frac{x}{4}=-8$

$x=-\text{32}$

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