###### Example 3.6.10

Solve for \(a\) in \(\frac{2}{3}(a+1)-\frac{5}{6}=\frac{2}{3}a\text{.}\)

To solve this equation for \(a\text{,}\) we'll want to recall the technique of multiplying each side of the equation by the LCD of all fractions. Here, this means that we will multiply each side by \(6\) as our first step. After that, we'll be able to simplify each side of the equation and continue solving for \(a\text{:}\)

\begin{align*}
\frac{2}{3}(a+1)-\frac{5}{6}\amp=\frac{2}{3}a\\
\multiplyleft{6}\left(\frac{2}{3}(a+1)-\frac{5}{6}\right)\amp=\multiplyleft{6}\frac{2}{3}a\\
\multiplyleft{6}\frac{2}{3}(a+1)-\multiplyleft{6}\frac{5}{6}\amp=\multiplyleft{6}\frac{2}{3}a\\
4(a+1)-5\amp=4a\\
4a+4-5\amp=4a\\
4a-1\amp=4a\\
4a-1\subtractright{4a}\amp=4a\subtractright{4a}\\
-1\amp=0
\end{align*}

The statement \(-1=0\) is false, so the equation has no solution. We can write the empty set as: \(\emptyset\text{.}\)