**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)#