How do you solve #x-y=-2, 2x+y=8# using graphing?

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Answer 1 · 2017-08-17T13:36:58

See a solution process below:

First, we can graph the first line by find two points on the line, plotting them and drawing a line through them:

For #x = 0#: #0 - y = -2#

#-y = -2#

#color(red)(-1) xx -y = color(red)(-1) xx -2#

#y = 2# or #(0, 2)#

For #y = 0#: #x - 0 = -2#

#x = -2# or #(-2, 0)#

graph{(x^2+(y-2)^2-0.125)((x+2)^2+y^2-0.125)(x-y+2)=0 [-20,20,-10,10]}

We can now do the same thing for the second equation:

#2x + y = 8#

For #x = 0#: #(2 * 0) + y = 8#

#0 + y = 8#

#y = 8# or #(0, 8)#

For #y = 0#: #2x + 0 = 8#

#2x = 8#

#(2x)/color(red)(2) = 8/color(red)(2)#

#x = 4# or #(4, 0)#

graph{(x^2+(y-8)^2-0.125)((x-4)^2+y^2-0.125)(2x+y-8)(x^2+(y-2)^2-0.125)((x+2)^2+y^2-0.125)(x-y+2)=0 [-20,20,-10,10]}

We can now see where the two lines cross:

graph{(2x+y-8)(x-y+2)((x-2)^2+(y-4)^2-0.05)=0 [-5,15,-5,5]}

They cross at #x = 2# and #y = 4# or #(2, 4)#