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

May 30, 2017

Solution is $x = 4$ and $y = 5$.

Explanation:

We select a few points satisfying the first equation i.e. $x + y = 9$. Let these be $\left(- 3 , 12\right)$, $\left(0 , 9\right)$ and $\left(5 , 4\right)$. Joining them gives the graph as follows:
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 $x - y = - 1$, let some of the points satisfying the equation be $\left(- 6 , - 5\right)$, $\left(- 1 , 0\right)$ and $\left(5 , 6\right)$ and graph appears as follows.
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 $\left(4 , 5\right)$ asshown below.
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 $x = 4$ and $y = 5$.

Note $-$ In case you find that lines are parallel and do not intersect, we have #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.