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

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

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

Check: When ${\displaystyle y=9}$

${\displaystyle 6y=\text{54}}$

becomes
,
a true statement.

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

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

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

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

${\displaystyle x=-\text{54}}$

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

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

becomes

a true statement.

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

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

Method 1: Use of dividing out common factors.

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

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

${\displaystyle \cancel{7}\cdot \frac{3a}{\cancel{7}}=\text{42}}$

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

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

${\displaystyle \frac{\cancel{3}a}{\cancel{3}}=\text{14}}$

${\displaystyle a=\text{14}}$

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

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

becomes
,
a true statement.

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

Method 2: Use of reciprocals

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

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

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

${\displaystyle \frac{\stackrel{1}{\cancel{7}}}{\underset{1}{\cancel{3}}}\cdot \frac{\stackrel{1}{\cancel{3}a}}{\underset{1}{\cancel{7}}}=\frac{7}{\underset{1}{\cancel{3}}}\cdot \frac{\stackrel{2}{\cancel{6}}}{1}}$

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

Notice that we get the same solution using either method.

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

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

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

${\displaystyle x=-3}$

Check: When ${\displaystyle x=-3}$ ,

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

becomes
,
a true statement.

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

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

Check: When ${\displaystyle x=7}$ ,

${\displaystyle -x=7}$

becomes

The solution to ${\displaystyle -x=7}$ is ${\displaystyle x=-7}$ .