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#