How do you graph #-x+y=9" and "-8x+8y=-8#?

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Answer 1 · 2018-02-03T12:17:08

See a solution process below:

To graph these two lines we need to solve for two points for each equation, plot the two points and then draw a line through the two points.

Equation 1: #-x + y = 9#

First Point: For #x = 0#

#-0 + y = 9#

#0 + y = 9#

#y = 9# or #(0, 9)#

Second Point: For #y = 0#

#-x + 0 = 9#

#-x = 9#

#-x xx color(red)(-1) = 9 xx color(red)(-1)#

#x = -9# or #(-9, 0)#

We can next plot the two points on the coordinate plane and draw a line through the two points:

graph{(-x+y-9)(x^2+(y-9)^2-0.125)((x+9)^2+y^2-0.125)=0 [-20, 20, -10, 10]}

Equation 2: #-8x + 8y = -8#

First Point: For #x = 0#

#(-8 xx 0) + 8y = -8#

#0 + 8y = -8#

#8y = -8#

#(8y)/color(red)(8) = -8/color(red)(8)#

#y = -1# or #(0, -1)#

Second Point: For #y = 0#

#-8x + (8 xx 0) = -8#

#-8x + 0 = -8#

#-8x = -8#

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

#x = 1# or #(1, 0)#

We can next plot the two points for the second equation on the coordinate plane and draw a line through the two points:

graph{(-8x+8y+8)(-x+y-9)(x^2+(y+1)^2-0.125)((x-1)^2+y^2-0.125)=0 [-20, 20, -10, 10]}