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)#
