How do you solve the following system: #3x-y=12, 2x-y/2=4 #?

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Answer 1 · 2016-02-10T10:55:00

#(-4,-24)#

#3x-y=12# -----(1)

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

multiply #eq^n # (2) by 2 we get,

#4x-y=8#----------(3)

subtract equation (1) from equation (2)

(#4x-y=8) - (3x-y=12)#

#4x-3x-y+y=8-12#

#x=8-12#

#x=-4#

Substitute in equation (1)

#3x-y=12#

#3(-4) -y =12#

#-12-y = 12#

#-12-12 = y#

#y=-24#