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

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Answer 1 · 2017-05-01T01:49:05

See the solution process below:

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

#x - y = 3#

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

#x - 0 = 3 + y#

#x = 3 + y#

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

#2x + 2y = 2# becomes:

#2(3 + y) + 2y = 2#

#(2 * 3) + (2 * y) + 2y = 2#

#6 + 2y + 2y = 2#

#6 + (2 + 2)y = 2#

#6 + 4y = 2#

#-color(red)(6) + 6 + 4y = -color(red)(6) + 2#

#0 + 4y = -4#

#4y = -4#

#(4y)/color(red)(4) = -4/color(red)(4)#

#(color(red)(cancel(color(black)(4)))y)/cancel(color(red)(4)) = -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 + y# becomes:

#x = 3 + (-1)#

#x = 3 - 1#

#x = 2#

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