How do you graph #y=2x-3# and #y=-3x+2#?

1 Answer
smendyka
Nov 18, 2017

See a solution process below:

Explanation:

First, solve for two points which solve the first equation and plot these points:

First Point: For #x = 0#

#y = (2 xx 0) - 3#

#y = 0 - 3#

#y = -3# or #(0, -3)#

Second Point: For #x = 2#

#y = (2 xx 2) - 3#

#y = 4 - 3#

#y = 1# or #(2, 1)#

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

graph{(x^2+(y+3)^2-0.025)((x-2)^2+(y-1)^2-0.025)(y-2x+3)=0 [-10, 10, -5, 5]}

Now, follow the same process for the other equation:

First Point: For #x = 0#

#y = (-3 xx 0) + 2#

#y = 0 + 2#

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

Second Point: For #x = 2#

#y = (-3 xx 2) + 2#

#y = -6 + 2#

#y = -4# or #(2, -4)#

We can then plot the two points for the second equation on the coordinate plane and draw a straight line through the two points to graph the second line:

graph{(x^2+(y-2)^2-0.025)((x-2)^2+(y+4)^2-0.025)(y+3x-2)(y-2x+3)=0 [-10, 10, -5, 5]}

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