How do you solve # x+5y=-3# and #3x-2y=8# using substitution?

1 Answer
Feb 12, 2017

Answer:

See the entire simplification process below:

Explanation:

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

#x + 5y = -3#

#x + 5y - color(red)(5y) = -3 - color(red)(5y)#

#x + 0 = -3 - 5y#

#x = -3 - 5y#

Step 2) Substitute #-3 - 5y# for #x# in the second equation and solve for #y#:

#3x - 2y = 8# becomes:

#3(-3 - 5y) - 2y = 8#

#(3 xx -3) - (3 xx 5y) - 2y = 8#

#-9 - 15y - 2y = 8#

#-9 - 17y = 8#

#color(red)(9) - 9 - 17y = color(red)(9) + 8#

#0 - 17y = 17#

#-17y = 17#

#(-17y)/color(red)(-17) = 17/color(red)(-17)#

#(color(red)(cancel(color(black)(-17)))y)/cancel(color(red)(-17)) = -1#

#y = -1#

Step 3) Substitute #-1# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = -3 - 5y# becomes:

#x = -3 - (5 xx -1)#

#x = -3 - (-5)#

#x = -3 + 5#

#x = 2#

The solution is: #x = 2# and #y = -1# or #(2, -1)#