How do you solve the system of equations #(2x)/y=12# and #8(y+2)=x#?

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Answer 1 · 2016-10-12T20:36:59

#x=-48#

#y=-8#

#(2x)/y=12 " (1)"#

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

#"rearrange the equation (1)"#

#cancel(2)x=cancel(12)y#

#x=color(green)(6y)#

#"now use the equation (2)"#

#8(y+2)=color(green)(6y)#

#8y+16=6y#

#8y-6y=-16#

#2y=-16#

#y=-16/2#

#y=-8#

#"use (1)"#

#(2x)/-8=12#

#cancel(2)x=-cancel(8)*12#

#x=-48#