How do you solve the system of equations $(4 x - 2 y = 12$ $(4x-2y=12$ and $3 x + 2 y = 2$ $3x+2y=2$ by elimination?

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Answer 1 · 2018-04-11T16:35:54

$x = 2$ $x=2$ , $y = - 2$ $y=-2$

From second equation, $2 y = 2 - 3 x$ $2y=2-3x$ . Hence,

$4 x - (2 - 3 x) = 12$ $4x-(2-3x)=12$

$4 x + 3 x - 2 = 12$ $4x+3x-2=12$

$7 x - 2 = 12$ $7x-2=12$

$7 x = 14$ $7x=14$ , so $x = 2$ $x=2$

Thus,

$2 y = 2 - 3 \cdot 2 = - 4$ $2y=2-3*2=-4$ or $y = - 2$ $y=-2$

How do you solve the system of equations (4x−2y=12(4x-2y=12 and 3x+2y=23x+2y=2 by elimination?