How do you solve the system #x+y=4# and #x-y=2# by graphing?

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Answer 1 · 2017-08-03T04:01:19

See a solution process below:

First, draw the line for the first equation using two points:

#x = 0#; then #0 + y = 4# or #y = 4# giving: #(0, 4)#

#y = 0#; then #x + 0 = 4# or #x = 4# giving: #(4, 0)#

graph{(x+y- 4)((x-4)^2+(y)^2-0.5)((x)^2+(y-4)^2-0.5)=0 [-40, 40, -20, 20]}

Next, draw the line for the second equation using two points:

#x = 0#; then #0 - y = 2# or #-y = 2# or #y = -2# giving #(0, -2)#

#y = 0#; then #x - 0 = 2# or #x = 2# or #(2, 0)#

graph{(x - y - 2)(x+y- 4)((x-4)^2+(y)^2-0.025)((x)^2+(y-4)^2-0.025)((x)^2+(y+2)^2-0.025)((x-2)^2+(y)^2-0.025)=0 [-10, 10, -5, 5]}

The lines are seen to intersect on the #x# axis at: #3#

The lines are seen to intersect on the #y# axis at: #1#

Therefore the solution is: #(3, 1)#