How do you solve the system of linear equations #4x+3y= -2# and #2x-2y= -8#?

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Answer 1 · 2018-07-30T05:15:22

#x=-2 and y=2#

Here ,

#color(red)(4x+3y=-2.....to(1)#

#color(red)(2x-2y=-8.....to(2)#

Dividing both sides of # eqn.(2)# by #2#

#color(red)(x-y=-4#

#=>color(red)(x=y-4.....to(3)#

Subst. value of #x# into # (1)# we get

#4(y-4)+3y=-2#

#=>4y-16+3y=-2#

#=>4y+3y=16-2#

#=>7y=14#

#=>color(blue)(y=2#

Subst. #y=2 # into #(3)#

#color(blue)(x=2-4=-2#

Hence, #x=-2 and y=2#

Answer 2 · 2018-07-30T11:31:38

#x=-2# and #y=2#

#2*(2x-2y)-(4x+3y)=2*(-8)-(-2)#

#4x-4y-4x-3y=-16+2#

#-7y=-14#

Hence #y=(-14)/(-7)=2#

So,

#2x-2*2=-8#

#2x-4=-8#

#2x=-4#

Consequently, #x=(-4)/2=-2#