How do you solve the following system?: $8 x + y = 12, 7 x - 5 y = - 4$ $8x +y =12, 7x -5y = -4$

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Answer 1 · 2018-04-13T10:06:19

$x = \frac{56}{47}$ $x=56/47$

So I first re write one equation in this case so we know what y is equal to. I rewrote:

$8 x + y = 12$ $8x+y=12$

So we get:

$y = 12 - 8 x$ $y=12-8x$

Then we take that and put it into the other equation so we get:

$7 x - 5 (12 - 8 x) = - 4$ $7x-5(12-8x)=-4$

Which is:

$7 x - 60 + 40 x = - 4$ $7x-60+40x=-4$

Simplified:

$7 x + 40 x = 56$ $7x+40x=56$

$47 x = 56$ $47x=56$

$x = \frac{56}{47}$ $x=56/47$

How do you solve the following system?: 8x+y=12,7x−5y=−48x +y =12, 7x -5y = -4