How do you solve the following system: $2x - 3y = 10 , 2x + 3y = 12$ ?

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Answer 1 · 2015-12-14T21:25:16

$color(white)(xx)x=11/2$ , and $y=1/3$

$color(white)(xx)2x-3y=10,color(white)(x)2x+3y=12$

$color(white)(xx)2x-3y=10<=>3y=2x-10$
$color(white)(xx)2x+3y=12<=>3y=-2x+12$

$=>2x-10=y$
$=>2x-10=color(red)(-2x+12)$

Add $color(red)(+2x+10)$ to both side:
$color(white)(xx)2x-10color(red)(+2x+10)=-2x+12color(red)(+2x+10)$
$=>4x=22$

Multiply both sides by $color(red)(1/4)$ :
$color(white)(xx)color(red)(1/4xx)4x=color(red)(1/4xx)22$
$=>x=11/2$

$color(white)(xx)3y=2x-10$
$=>3y=2xxcolor(red)(11/2)-10$
$=>3y=1$

$=>color(red)(1/3xx)3y=color(red)(1/3xx)1$
$=>y=1/3$

How do you solve the following system: 2x - 3y = 10 , 2x + 3y = 122x - 3y = 10 , 2x + 3y = 12?