Question

solve the following system of equations graphically on the set of axes below.
$y = -\frac{2}{3}x - 5$
$x - 3y = -3$
plot two lines by clicking the graph.
click a line to delete it.

Explanation:

Step1: Rewrite 2nd equation to slope-intercept

Rearrange $x-3y=-3$ to $y=mx+b$:
$-3y = -x -3$
$y = \frac{1}{3}x + 1$

Step2: Find points for first line

For $y=-\frac{2}{3}x-5$:

• When $x=0$, $y=-5$ → $(0, -5)$
• When $x=3$, $y=-\frac{2}{3}(3)-5=-7$ → $(3, -7)$

Step3: Find points for second line

For $y=\frac{1}{3}x+1$:

• When $x=0$, $y=1$ → $(0, 1)$
• When $x=3$, $y=\frac{1}{3}(3)+1=2$ → $(3, 2)$

Step4: Find intersection algebraically

Set equations equal:
$-\frac{2}{3}x -5 = \frac{1}{3}x +1$
Multiply by 3: $-2x -15 = x +3$
$-3x = 18$
$x=-6$
Substitute $x=-6$ into $y=\frac{1}{3}x+1$:
$y=\frac{1}{3}(-6)+1=-2+1=-1$

Answer:

The solution (intersection point) is $\boldsymbol{(-6, -1)}$
To plot:

1. For $y=-\frac{2}{3}x-5$: plot $(0, -5)$ and $(3, -7)$, draw the line through them.
2. For $y=\frac{1}{3}x+1$: plot $(0, 1)$ and $(3, 2)$, draw the line through them.

The lines cross at $(-6, -1)$.