How do you solve the system by substitution $y = 4x + 12$ and $y = 2x + 8$ ?

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Answer 1 · 2016-01-01T13:03:16

$x=-2, y=4$

$y=4x+12$

$y-4x=12$ --- $(1)$

$y=2x+8$ --- $(2)$

Subsitute (2) into (1)

$2x+8-4x=12$

$-2x=-8+12$

$-2x=4$

$x=-2$

Subsitute x=-2 into (2)

$y=2(-2)+8$

$y=-4+8$

$y=4$

$x=-2, y=4$

How do you solve the system by substitution y = 4x + 12y = 4x + 12 and y = 2x + 8y = 2x + 8?