Question

Solve the following linear system by elimination

Answer 1

so hmmm let's notice the denominators in each, so hmmm the 1st equation has an LCD of 6, so let's multiply both sides by that, and the 2nd equation has an LCD of 12, so let's multiply it by that to do away with the denominators, oddly enough, it gives us an elimination of a variable, so let's do so

`\begin{array}{llll} \cfrac{1}{2}x-\cfrac{2}{3}y=6 \\\\\\ \cfrac{1}{4}x+\cfrac{1}{3}y=-1 \end{array}\hspace{5em} \begin{array}{llll} 6\left( \cfrac{1}{2}x-\cfrac{2}{3}y=6 \right) \\\\\\ 12\left( \cfrac{1}{4}x+\cfrac{1}{3}y=-1 \right) \end{array}\hspace{5em} \begin{array}{llll} ~~ 3x-4y&=36 \\\\\\\\\\ -3x-4y&=12\\\cline{1-2} ~\hfill -8y&=48 \end{array} \\\\[-0.35em] ~\dotfill`

`-8y=48\implies y=\cfrac{48}{-8}\implies y=-6 \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{3x-4y=36}\implies 3x-4(-6)=36\implies 3x+24=36 \\\\\\ 3x=12\implies x=\cfrac{12}{3}\implies x=4 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\Large \begin{array}{llll} (4~~,~-6) \end{array}}~\hfill`