Step 1) Because the second equation is solved for #y# we can substitute #(-2x + 1)# for #y# in the first equation and solve for #x#:
#-3x - 4y = 5# becomes:
#-3x - 4(-2x + 1) = 5#
#-3x + (4 * 2x) - (4 + 1) = 5#
#-3x + 8x - 4 = 5#
#(-3 + 8)x - 4 = 5#
#5x - 4 = 5#
#5x - 4 + color(red)(4) = 5 + color(red)(4)#
#5x - 0 = 9#
#5x = 9#
#(5x)/color(red)(5) = 9/color(red)(5)#
#(color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5)) = 9/5#
#x = 9/5#
Step 2) Substitute #9/5# for #x# in the second equation and calculate #y#:
#y = -2x + 1# becomes:
#y = (-2 xx 9/5) + 1#
#y = -18/5 + 1#
#y = -18/5 + (5/5 xx 1)#
#y = -18/5 + 5/5#
#y = (-18 + 5)/5#
#y = -13/5#
The solution is: #x = 9/5# and #y = -13/5# or #(9/5, -13/5)#
