How do you solve the system of equations #3x+2y = 14# and #y = x+2# by substitution?

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Answer 1 · 2017-08-12T21:52:29

See a solution process below:

Step 1) Because the second equation is already solved for #y# we can substitute #(x + 2)# for #y# in the first equation and solve for #x#:

#3x + 2y = 14# becomes:

#3x + 2(x + 2) = 14#

#3x + (2 xx x) + (2 xx 2) = 14#

#3x + 2x + 4 = 14#

#(3 + 2)x + 4 = 14#

#5x + 4 = 14#

#5x + 4 - color(red)(4) = 14 - color(red)(4)#

#5x + 0 = 10#

#5x = 10#

#(5x)/color(red)(5) = 10/color(red)(5)#

#(color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5)) = 2#

#x = 2#

Step 2) Substitute #2# for #x# in the second solution and solve for #y#:

#y = x + 2# becomes:

#y = 2 + 2#

#y = 4#

The Solution Is: #x = 2# and #y = 4# or #(2, 4)#