How do you solve the system $3 x + 2 y = 2, - 2 x + y = 8$ $3x+2y=2, -2x+y=8$ ?

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Answer 1 · 2016-12-26T09:03:12

$x = - 2$ $x=-2$
$y = 4$ $y=4$

Since, by the second equation, it is

$y = 8 + 2 x$ $y=8+2x$ $(1)$ $color(red)((1))$

you can put it in the first equation and get:

$3 x + 2 (8 + 2 x) = 2$ $3x+2(8+2x)=2$

Then you multiply and get:

$3 x + 16 + 4 x = 2$ $3x+16+4x=2$

that's

$7 x = 2 - 16$ $7x=2-16$

$7 x = - 14$ $7x=-14$

$x = - 2$ $x=-2$

Then you put it in $(1)$ $color(red)((1))$ and get:

$y = 8 + 2 (- 2) = 8 - 4 = 4$ $y=8+2(-2)=8-4=4$

How do you solve the system 3x+2y=2, -2x+y=83x+2y=2,−2x+y=83x+2y=2, -2x+y=8?