How do you solve #y=x-4# and #y=-x+2#?

1 Answer
Feb 11, 2017

Answer:

See the entire solution process below:

Explanation:

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)#