How do you solve #4x+5y=21# and #-3x+ y =27# using substitution?

1 Answer
Mar 29, 2016

Answer:

#color(brown)("Only demonstrating a different approach")#
This uses substitution to find #y# if you plug the value of #x# into one of the equations. However I should think they were looking for the method the others used.

Explanation:

Given

#color(white)(.....)4x+5y=21 #...............................(1)
#color(white)(.)-3x+y=27#.................................(2)

Multiply equation (2) by 5 giving:

#color(white)(.......)4x+5y=21 #...............................(1)
#underline(color(white)(.)-15x+5y=135)" "...............(2_a)#
#color(white)(......)19x+color(white)(.)0=-114" "-># Subtract equation #(2_a) # from (1)

#=> x=-114/19 = -6#

Then solve as the others did