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

1 Answer
Shwetank Mauria
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 #(-3,12)#, #(0,9)# and #(5,4)#. 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 #(-6,-5)#, #(-1,0)# and #(5,6)# 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 #(4,5)# 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.

Related questions
See all questions in Graphs of Linear Systems
Impact of this question
4998 views around the world
You can reuse this answer
Creative Commons License