Step 1) Because the first equation is already solved for yy, substitute x - 4x−4 for yy in the second equation and solve for xx:
y = -x + 2y=−x+2 becomes:
x - 4 = -x + 2x−4=−x+2
x - 4 + color(red)(4) + color(blue)(x) = -x + 2 + color(red)(4) + color(blue)(x)x−4+4+x=−x+2+4+x
x + color(blue)(x) - 4 + color(red)(4) = -x + color(blue)(x) + 2 + color(red)(4)x+x−4+4=−x+x+2+4
1x + color(blue)(1x) - 0 = 0 + 61x+1x−0=0+6
2x = 62x=6
(2x)/color(red)(2) = 6/color(red)(2)2x2=62
(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)