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

##### 1 Answer
Aug 3, 2017

See a solution process below:

#### Explanation:

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

$x = 0$; then $0 + y = 4$ or $y = 4$ giving: $\left(0 , 4\right)$

$y = 0$; then $x + 0 = 4$ or $x = 4$ giving: $\left(4 , 0\right)$

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 $\left(0 , - 2\right)$

$y = 0$; then $x - 0 = 2$ or $x = 2$ or $\left(2 , 0\right)$

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: $\left(3 , 1\right)$