How do you solve the following system?: $2x + 7y = -8, -x-8y=32$

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Answer 1 · 2016-04-25T16:57:12

$x=160/9$ and $y=-56/9$ .

The equations are $2x+7y=-8$ --(A) and $-x-8y=32$ -- (B)

Now multiplying (B) by $2$ and adding to (A) cancels out $x$ , and we get

$7y-16y=-8+64$ or $-9y=56$ or $y=-56/9$ . Now putting this in (B), we get

$-x-8xx-56/9=32$ or $-x+448/9=32$ or $-x=32-448/9=288/9-448/9=-160/9$ or $x=160/9$

How do you solve the following system?: 2x + 7y = -8, -x-8y=32 2x + 7y = -8, -x-8y=32