How do you solve the system #x-5y=36# and #2x+y=-16# using substitution?

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Answer 1 · 2017-05-31T12:51:49

See a solution process below:

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

#x - 5y = 36#

#x - 5y + color(red)(5y) = 36 + color(red)(5y)#

#x - 0 = 36 + 5y#

#x = 36 + 5y#

Step 2) Substitute #(36 + 5y)# for #x# in the second equation and solve for #y#:

#2x + y = -16# becomes:

#2(36 + 5y) + y = -16#

#(2 xx 36) + (2 xx 5y) + y = -16#

#72 + 10y + y = -16#

#72 + 10y + 1y = -16#

#72 + (10 + 1)y = -16#

#72 + 11y = -16#

#-color(red)(72) + 72 + 11y = -color(red)(72) - 16#

#0 + 11y = -88#

#11y = -88#

#(11y)/color(red)(11) = -88/color(red)(11)#

#(color(red)(cancel(color(black)(11)))y)/cancel(color(red)(11)) = -8#

#y = -8#

Step 3) Substitute #-8# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = 36 + 5y# becomes:

#x = 36 + (5 xx -8)#

#x = 36 + (-40)#

#x = -4#

The solution is: #x = -4# and #y = -8# or #(-4, -8)#