Step 1) Solve the second equation for #x#:
#z - 2y + color(red)(2y) = -3 + color(red)(2y)#
#z - 0 = -3 + 2y#
#z = -3 + 2y#
Step 2) Substitute #(-3 + 2y)# for #x# in the first equation and solve for #y#:
#5x - 4y = 9# becomes:
#5(-3 + 2y) - 4y = 9#
#(5 xx -3) + (5 xx 2y) - 4y = 9#
#-15 + 10y - 4y = 9#
#-15 + (10 - 4)y = 9#
#-15 + 6y = 9#
#color(red)(15) - 15 + 6y = color(red)(15) + 9#
#0 + 6y = 24#
#6y = 24#
#(6y)/color(red)(6) = 24/color(red)(6)#
#(color(red)(cancel(color(black)(6)))y)/cancel(color(red)(6)) = 4#
#y = 4#
Step 3) Substitute #4# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:
#x = -3 + 2y# becomes:
#x = -3 + (2 xx 4)#
#x = -3 + 8#
#x = 5#
The solution is: #x = 5# and #y = 4# or #(5, 4)#