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Solve by elimination 3x-2y=2 3x+4y=50

User Dozer
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Given the system of equations :


\begin{gathered} 3x-2y=2 \\ 3x+4y=50 \end{gathered}

Subtract the first equation from the second equation to eliminate x

Then, solve to find y


\begin{gathered} (3x+4y)-(3x-2y)=50-2 \\ (3x-3x)+(4y-(-2y))=50-2 \\ 6y=48 \\ \\ y=(48)/(6)=8 \end{gathered}

Substitute with y = 8 at the first equation to find x :


\begin{gathered} 3x-2\cdot8=2 \\ 3x-16=2 \\ 3x=2+16 \\ 3x=18 \\ \\ x=(18)/(3)=6 \end{gathered}

So, the answer of the system :


\begin{gathered} x=6 \\ y=8 \\ (x,y)=(6,8) \end{gathered}

User Nhuluseda
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