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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.
The multiplication/division property of equality can be used to undo an association with a number that multiplies or divides the variable.
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
$\u2013x$ 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 $x=-7$ .
Use the multiplication/division property of equality to solve each equation. Be sure to check each solution.
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