# How do you solve #x+y=9#, #x-y=-1# by graphing and classify the system?

##### 1 Answer

#### Answer:

Solution is

#### Explanation:

We select a few points satisfying the first equation i.e.

graph{(x+y-9)((x+3)^2+(y-12)^2-0.03)(x^2+(y-9)^2-0.03)((x-5)^2+(y-4)^2-0.03)=0 [-9.57, 10.43, 2.64, 12.64]}

Similarly for second equation

graph{(x-y+1)((x+6)^2+(y+5)^2-0.05)((x+1)^2+y^2-0.05)((x-5)^2+(y-6)^2-0.05)=0 [-21.28, 18.72, -9, 11]}

If we draw the two graphs on te same Cartesian plane they intersect each other at

graph{(x+y-9)(x-y+1)((x-4)^2+(y-5)^2-0.05)=0 [-18.11, 21.89, -5.32, 14.68]}

Hence, solution is

**Note** **no solution**. Some times two lines may coincide.. This means the two lines are essentially same. In such cases you wil find that the one line is just a multiple of anther line.