Example 3.5.4

Solve \(\frac{x}{8} = \frac{15}{12}\) for \(x\text{.}\)

Instead of finding the LCD of the two fractions, we'll simply multiply both sides of the equation by \(8\) and by \(12\text{.}\) This will still have the effect of canceling the denominators on both sides of the equation.

\begin{align*} \frac{x}{8} \amp= \frac{15}{12}\\ \highlight{12\cdot8\cdot}\frac{x}{8} \amp= \frac{15}{12}\highlight{\cdot12\cdot8}\\ 12\cdot\cancelhighlight{8}\cdot\frac{x}{\cancelhighlight{8}} \amp= \frac{15}{\cancelhighlight{12}}\cdot\cancelhighlight{12}\cdot8\\ 12\cdot x \amp= 15\cdot 8\\ 12x \amp= 120 \\ \divideunder{12x}{12} \amp= \divideunder{120}{12} \\ x \amp= 10 \end{align*}

Our work indicates \(10\) is the solution. We can check this as we would for any equation, by substituting \(10\) for \(x\) and verifying we obtain a true statement:

\begin{align*} \frac{10}{8} \amp\stackrel{?}{=} \frac{15}{12}\\ \frac{5}{4} \amp\stackrel{\checkmark}{=} \frac{5}{4} \end{align*}

Since both fractions reduce to \(\frac{5}{4}\text{,}\) we know the solution to the equation \(\frac{x}{8} = \frac{15}{12}\) is \(10\) and the solution set is \(\left\{10\right\}\text{.}\)

in-context