How do you solve #y=x+4#, #y=-2x+1# by graphing?

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Answer 1 · 2017-08-13T18:06:55

See a solution process below:

First graph the equation #y = x + 4# by plotting two points on the line and then drawing a line through the two points:

Graph Equation 1:

First graph the equation #y = x + 4# by plotting two points on the line and then drawing a line through the two points:

For #x = 0#; #y = 0 + 4 = 4# or #(0, 4)#

For #x = 4#: #y = 4 + 4 = 8# or #(4, 8)#

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

Graph Equation 2:

We do the same for the second equation:

For #x = 0#: #y = (-2 * 0) + 1 = 0 + 1 = 1# or #(0, 1)#

For #x = -3#: #y = (-2 * -3) + 1 = 6 + 1 = 7# or #(-3, 7)#

graph{((x+3)^2+(y-7)^2-0.125)(y+2x-1)(x^2+(y-1)^2- 0.125)(x^2+(y-4)^2-0.125)((x-4)^2+(y-8)^2-0.125)(y-x-4)=0 [-20, 20, -10, 10]}

The two lines intersect at: #(-1, 3)#

graph{((x+1)^2+(y-3)^2-0.075)(y+2x-1)(y-x-4)=0 [-10, 10, -5, 5]}