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

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Answer 1 · 2016-01-01T13:03:16

$x = - 2, y = 4$ $x=-2, y=4$

$y = 4 x + 12$ $y=4x+12$

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

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

Subsitute (2) into (1)

$2 x + 8 - 4 x = 12$ $2x+8-4x=12$

$- 2 x = - 8 + 12$ $-2x=-8+12$

$- 2 x = 4$ $-2x=4$

$x = - 2$ $x=-2$

Subsitute x=-2 into (2)

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

$y = - 4 + 8$ $y=-4+8$

$y = 4$ $y=4$

$x = - 2, y = 4$ $x=-2, y=4$

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