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

Oct 17, 2017

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 $x = 0$

$\left(- 3 \cdot 0\right) + y = - 3$

$0 + y = - 3$

$y = - 3$ or $\left(0 , - 3\right)$

Second Point: For $x = 1$

$\left(- 3 \cdot 1\right) + y = - 3$

$- 3 + y = - 3$

$\textcolor{red}{3} - 3 + y = \textcolor{red}{3} - 3$

$0 + y = 0$

$y = 0$ or $\left(1 , 0\right)$

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 $\left(0 , - 3\right)$

Second Point: For $x = 1$

$y = 1 - 3$

$y = - 2$ or $\left(1 , - 2\right)$

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 $\left(0 , - 3\right)$