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

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Answer 1 · 2018-04-11T16:35:54

#x=2#, #y=-2#

From second equation, #2y=2-3x#. Hence,

#4x-(2-3x)=12#

#4x+3x-2=12#

#7x-2=12#

#7x=14#, so #x=2#

Thus,

#2y=2-3*2=-4# or #y=-2#