How do you solve the system of equations $10 x + 8 y = 2$ $10x + 8y = 2$ and $- 2 x - 4 y = 6$ $- 2x - 4y = 6$ ?

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Answer 1 · 2018-06-22T18:16:11

$x = \frac{7}{3} and y = - \frac{8}{3}$ $x=7/3 and y=-8/3$

From first equation: $y=(2-10x)/8=(cancel2(1-5x))/cancel8=(1-5x)/4$

Substitute to second equation: $-2x-cancel(4)(1-5x)/cancel(4)=6$

So: $-2x-1+5x= 6 -> x=7/3$

Then: $y=(2-10(7/3))/8=(2-70/3)*1/8=-cancel64/3*1/cancel8=-8/3$

Answer 2 · 2018-06-22T18:19:16

$x=7/3,y=-8/3$

Multiplying the second equation by $5$ and adding to the first we get

$-6y=16$
so

$y=-8/3$
so we get

$-x+16/3=3$

$x=7/3$

How do you solve the system of equations 10x + 8y = 210x+8y=210x + 8y = 2 and - 2x - 4y = 6−2x−4y=6- 2x - 4y = 6?