Step 1) Because the first equation is already solved for #y#, substitute #x - 4# for #y# in the second equation and solve for #x#:
#y = -x + 2# becomes:
#x - 4 = -x + 2#
#x - 4 + color(red)(4) + color(blue)(x) = -x + 2 + color(red)(4) + color(blue)(x)#
#x + color(blue)(x) - 4 + color(red)(4) = -x + color(blue)(x) + 2 + color(red)(4)#
#1x + color(blue)(1x) - 0 = 0 + 6#
#2x = 6#
#(2x)/color(red)(2) = 6/color(red)(2)#
#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 3#
#x = 3#
Step 2) Substitute #3# for #x# in the first equation and calculate #y#:
#y = x - 4# becomes:
#y = 3 - 4#
#y = -1#
The solution is: #x = 3# and #y = -1# or #(3, -1)#
