# Solve the following problem by elimination method:2x=5y+4 and 3x-2y+16=0

$2x=5y+4\phantom{\rule{0ex}{0ex}}So,x=\frac{5y+4}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3x-2y+16=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3\left(\frac{5y+4}{2}\right)-2y+16=0\phantom{\rule{0ex}{0ex}}\frac{15y+12}{2}-2y+16=0\phantom{\rule{0ex}{0ex}}15y+12-4y+32=0\phantom{\rule{0ex}{0ex}}11y+44=0\phantom{\rule{0ex}{0ex}}y=-4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}So,x=\frac{5y+4}{2}=\frac{5(-4)+4}{2}=-8\phantom{\rule{0ex}{0ex}}$

Regards

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