How do you solve #3x+4y=-4# and #x+2y=2# using substitution?

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Answer 1 · 2017-01-24T14:48:49

See the entire solution process below:

Step 1) Solve the second equation for #x#:

#x + 2y = 2#

#x + 2y - 2y = 2 - 2y#

#x + 0 = 2 - 2y#

#x = 2 - 2y#

Step 2) Substitute #color(red)(2 - 2y)# for #x# in the first equation and solve for #y#:

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

#6 - 6y + 4y = -4#

#6 - 2y = -4#

#6 - 6 - 2y = -4 - 6#

#0 - 2y = -10#

#-2y = -10#

#(-2y)/color(red)(-2) = (-10)/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2)) = 5#

#y = 5#

Step 3) Substitute #color(red)(5)# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 2 - (2 xx color(red)(5))#

#x = 2 - 10#

#x = -8#

The solution is:

#x = -8# and #y = 5# or (-8, 5)