How do you solve the system of equations #6y - 4( x - 1) = - 24# and #3( y + 1) - x = - 10#?

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Answer 1 · 2018-07-03T21:15:10

#x=1, y=-4#

Given that

#6y-4(x-1)=-24#

#2x-3y=14\ .......(1)#

&

#3(y+1)-x=-10#

#x-3y=13\ .......(2)#

Subtracting (1) from (2), we get

#2x-3y-(x-3y)=14-13#

#x=1#

setting #x=1# in (1), we get

#2\cdot 1-3y=14#

#3y=-12#

#y=-4#