How do you solve #x=2y+7# and #3x−2y=3 # using substitution?

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Answer 1 · 2017-08-24T13:14:33

See a solution process below:

Step 1) Because the first equation is already solved for #x# we can substitute #(2y + 7)# for #x# in the second equation and solve for #y#:

#3x - 2y = 3# becomes:

#3(2y + 7) - 2y = 3#

#(3 * 2y) + (3 * 7) - 2y = 3#

#6y + 21 - 2y = 3#

#6y - 2y + 21 = 3#

#(6 - 2)y + 21 = 3#

#4y + 21 = 3#

#4y + 21 - color(red)(21) = 3 - color(red)(21)#

#4y + 0 = -18#

#4y = -18#

#(4y)/color(red)(4) = -18/color(red)(4)#

#(color(red)(cancel(color(black)(4)))y)/cancel(color(red)(4)) = -18/4#

#y = -9/2#

Step 2) Substitute #-9/2# for #y# in the first equation and calculate #x#:

#x = 2y + 7# becomes:

#x = (2 xx -9/2) + 7#

#x = (color(red)(cancel(color(black)(2))) xx -9/color(red)(cancel(color(black)(2)))) + 7#

#x = -9 + 7#

#x = -2#

The Solution Is: #x = -2# and #y = -9/2# or #(-2, -9/2)#