How do you solve the system $3 x + y = 12$ $3x+y=12$ and $y = - x - 2$ $y=-x-2$ using substitution?

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Answer 1 · 2017-09-29T15:23:34

$x = 7$ $x=7$
$y = 9$ $y=9$

From the second equation you know what y is equal to. You can substitute this into the first equation.

$3 x + (- x - 2) = 12$ $3x+(-x-2)=12$

$3 x - x - 2 = 12$ $3x-x-2=12$

$3 x - x = 12 + 2$ $3x-x=12+2$

$2 x = 14$ $2x=14$

$x = 7$ $x=7$

Now we can substitute our value of x into the second equation to find y.

$y = - (7) - 2$ $y=-(7)-2$

$y = - 7 - 2$ $y=-7-2$

$y = - 9$ $y=-9$

How do you solve the system 3x+y=123x+y=123x+y=12 and y=-x-2y=−x−2y=-x-2 using substitution?