###### Example3.6.10

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: $$\emptyset\text{.}$$

in-context