[tex]\bf \cfrac{4a}{3}-\cfrac{b}{4}=6
\qquad \qquad
\cfrac{5a}{6}+b=13
\\\\\\
\textit{let us remove the denominators off those folks}\\
\textit{by multiplying the first one by 12, both sides}\\
\textit{and the second one by 6, both sides, thus}
\\\\\\
12\left( \cfrac{4a}{3}-\cfrac{b}{4} \right)=12(6)\implies 16a-4b=72
\\\\\\
6\left( \cfrac{5a}{6}+b \right)=6(13)\implies 5a+6b=78[/tex]
[tex]\bf \textit{now, let's do the elimination}
\\\\
\begin{array}{llll}
16a-4b=72&\boxed{\times 3}\implies &48a-\underline{12b}=216\\\\
5a+6b=78&\boxed{\times 2}\implies &10a+\underline{12b}=156\\
&&--------\\
&&58a+0\quad=372
\end{array}[/tex]
solve for "a", once you get "a", plug it back into either equation to get "b"
