How do you solve #3x+4y=-11# and #7x-5y=3# using substitution?

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Answer 1 · 2016-07-17T03:28:18

You eliminate one variable (say #y#), ending up with an equation for only#x#, leading to the solution #x = -1, y=-2#

You start by rewriting the first equation as

#y = -1/4 (3x+11)#

Substituting this in the second equation gives

#7x +5/4(3x+11) = 3 #

or

#28x+15x+55 = 12#

Thus #43x = -43#, leading to #x=-1#., and

#y=-1/4 (3 xx (-1)+11) = -2#