Step 1) Because the second equation is already solved for #y#, we can substitute #(3x - 5)# for #y# in the first equation and solve for #x#:
#2x + 3y = 18# becomes:
#2x + 3(3x - 5) = 18#
#2x + (3 xx 3x) - (3 xx 5) = 18#
#2x + 9x - 15 = 18#
#(2 + 9)x - 15 = 18#
#11x - 15 = 18#
#11x - 15 + color(red)(15) = 18 + color(red)(15)#
#11x - 0 = 33#
#11x = 33#
#(11x)/color(re)(11) = 33/color(re)(11)#
#(color(red)(cancel(color(black)(11)))x)/cancel(color(re)(11)) = 3#
#x = 3#
Step 2) Substitute #3# for #x# in the second equation and solve for #y#:
#3x - 5 = y# becomes:
#(3 xx 3) - 5 = y#
#9 - 5 = y#
#4 = y#
#y = 4#
The Solution Is: #x = 3# and #y = 4# or #(3, 4)#
