How do you solve the system of equations #y= 3x + 12# and #2x - 2y = - 4#?

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Answer 1 · 2018-06-20T15:40:18

Substitution.

#y=3x+12 " " " "(1)#

#2x-2y =-4#

#-2y=-4-2x#

So

#y=(-4-2x)/-2#

#y= 2+x" " " "(2)#

Sub #(2)# into #(1)#

#3x+ 12= 2 +x#

#2x=-10#

#x=-5#

Sub #x=-5# into #(1)# (I would normally sub it into a given equation as I may have done a mistake in my new equation)

#y= 3(-5)+12#

#y=-3#

Therefore,

#y= -3 and x= -5#