How do you solve the system #-2x+3y=0# and #3x+3y=-15# by graphing?

1 Answer
smendyka
Aug 5, 2017

See a solution process below:

Explanation:

First, find two point on the line for the first equation.

Set #x = 0 : (-2 * 0) + 3y = 0#

#0 + 3y = 0#

#3y = 0#

#(3y)/color(red)(3) = 0/color(red)(3)#

#(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = 0#

#y = 0#

#(0, 0)#

Set #x = 3: (-2 * 3) + 3y = 0#

#-6 + 3y = 0#

#color(red)(6) - 6 + 3y = color(red)(6) + 0#

#0 + 3y = 6#

#3y = 6#

#(3y)/color(red)(3) = 6/color(red)(3)#

#(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = 2#

#y = 2#

#(3, 2)#

Now plot the two points and draw a line through the points:

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

Find two points on the line for the second equation.

Set #x = 0: (3 * 0) + 3y = -15#

#0 + 3y = -15#

#3y = -15#

#(3y)/color(red)(3) = -15/color(red)(3)#

#(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = -5#

#y = -5#

#(0, -5)#

Set #y = 0: 3x + (3 * 0) = -15#

#3x + 0 = -15#

#3x = -15#

#(3x)/color(red)(3) = -15/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = -5#

#x = -5#

#(-5, 0)#

Now, plot these two points and draw a line through them.

graph{((x+5)^2+y^2 - 0.125)(3x + 3y + 15)(x^2+(y+5)^2 - 0.125)(x^2 + y^2 - 0.125)((x-3)^2+(y - 2)^2 - 0.125)(-2x + 3y)=0 [-15, 15, -7.5, 7.5]}

The two lines intersect at: #(-3, -2)#

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