How do you solve this system of equations: $3x = 16+ 2y ; 5y = - 2x - 2$ ?

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Answer 1 · 2018-07-11T12:29:29

$x=4,y=-2$

From the second equation we get

$y=1/5(-2x-2)$
plugging this (for $y$ ) in the first equation

$3x=16+2(1/5(-2x-2))$
$3x=16+2/5(-2x-2)$
multiplying by $5$
$15x=80-4x-4$
so we get

$19x=76$
$x=4$

and

$y=1/5*(-8-2)=-10/5=-2$

How do you solve this system of equations: 3x = 16+ 2y ; 5y = - 2x - 23x = 16+ 2y ; 5y = - 2x - 2?