How do you solve the following system?: $x + 7 y = 3, 12 x + 3 y = - 2$ $x +7y = 3 , 12x + 3y = -2$

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Answer 1 · 2018-07-09T07:02:58

Multiply the first equation by 12

$12 x + 84 y = 36$ $12x+84y=36$

We have $12 x$ $12x$ in both equations so subtract the second equation from the first

$12 x + 84 y = 36$ $12x+84y=36$
$12 x + 3 y = - 2$ $12x+3y= -2$

$81 y = 38$ $81y=38$

$y = \frac{38}{81}$ $y=38/81$

Substitute this into equation two

$12 x + 3 \times \frac{38}{81} = - 2$ $12x+3xx38/81= -2$

$12 x + \frac{38}{27} = - 2$ $12x+38/27= -2$

$12 x = - \frac{92}{27}$ $12x= -92/27$

$x = - \frac{23}{81}$ $x= -23/81$

How do you solve the following system?: x+7y=3,12x+3y=−2x +7y = 3 , 12x + 3y = -2