How do you solve #x + y = 3 # and #3x + 4y = 8# using substitution?

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Answer 1 · 2016-12-06T21:36:46

#x = 4# and #y = -1#

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

#x - x + y = 3 - x#

#y = 3 - x#

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

#3x + 4(3 - x) = 8#

#3x + 12 - 4x = 8#

#12 - x = 8#

#12 - 8 - x + x = 8 - 8 + x#

#4 = x#

#x = 4#

Step 3) substitute #4# for #x# in the solution to the first equation in Step 1) and calculate #y#:

#y = 3 - 4#

#y = -1#