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

1 Answer
Oct 17, 2017

Answer:

See a solution process below:

Explanation:

We can first graph the line for the first equation. We can do this by solving for two points which solve the equation and plot these points and draw a straight line through the two points.:

First Point: For #x = 0#

#(-3 * 0) + y = -3#

#0 + y = -3#

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

Second Point: For #x = 1#

#(-3 * 1) + y = -3#

#-3 + y = -3#

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

#0 + y = 0#

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

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

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

We can follow this same process for the second equation.

First Point: For #x = 0#

#y = 0 - 3#

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

Second Point: For #x = 1#

#y = 1 - 3#

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

We can then plot the two points on the coordinate plane and draw the line:

graph{(y - x + 3)((x-1)^2+(y+2)^2-0.025)(-3x + y + 3)(x^2+(y+3)^2-0.05)((x-1)^2+y^2-0.025)=0 [-10, 10, -5, 5]}

We can see from the graph the two lines intersect at #(0, -3)#