Question

4a/3-b/4=6 5a/6+b=13 Solve linear system by elimination

Answer 1

$\bf \cfrac{4a}{3}-\cfrac{b}{4}=6 \qquad \qquad \cfrac{5a}{6}+b=13 \\\\\\ \textit{let us remove the denominators off those folks}\\ \textit{by multiplying the first one by 12, both sides}\\ \textit{and the second one by 6, both sides, thus} \\\\\\ 12\left( \cfrac{4a}{3}-\cfrac{b}{4} \right)=12(6)\implies 16a-4b=72 \\\\\\ 6\left( \cfrac{5a}{6}+b \right)=6(13)\implies 5a+6b=78$

$\bf \textit{now, let's do the elimination} \\\\ \begin{array}{llll} 16a-4b=72&\boxed{\times 3}\implies &48a-\underline{12b}=216\\\\ 5a+6b=78&\boxed{\times 2}\implies &10a+\underline{12b}=156\\ &&--------\\ &&58a+0\quad=372 \end{array}$

solve for "a", once you get "a", plug it back into either equation to get "b"