How do you solve #x-y=-2#, #2x+y=6# by graphing?

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Answer 1 · 2017-04-28T04:09:21

Solution is #(4/3,10/3)#

To solve such linear equations we should draw the graphs of these lines and point of intersection is the solution.

Let us draw graph of #x-y=2#. As some points on the line are

#(0,2)#, #(5,7)# and #(-8,-6)#, the graph appears as

graph{(x-y+2)(x^2+(y-2)^2-0.04)((x-5)^2+(y-7)^2-0.04)((x+8)^2+(y+6)^2-0.04)=0 [-20, 20, -10, 10]}

Similarly, some points on #2x+y=6# are #(0,6)#, #(4,-2)# and #(-1,8)# and graph appears as

graph{(2x+y-6)(x^2+(y-6)^2-0.04)((x-4)^2+(y+2)^2-0.04)((x+1)^2+(y-8)^2-0.04)=0 [-20, 20, -10, 10]}

The point of intersection is given by the graph below

graph{(2x+y-6)(x-y+2)=0 [-9.625, 10.375, -2.32, 7.68]}

i.e. Solution is #(4/3,10/3)#