Question

solve the following system of equations graphically on the set of axes below.
$y = -dfrac{2}{3}x - 3$
$y = dfrac{1}{3}x + 3$
plot two lines by clicking the graph.
click a line to delete it.

Explanation:

Step1: Analyze the first equation \( y = -\frac{2}{3}x - 3 \)

To graph this line, we can find two points. When \( x = 0 \), \( y = -3 \). When \( y = 0 \), \( 0 = -\frac{2}{3}x - 3 \), so \( \frac{2}{3}x = -3 \), \( x = -\frac{9}{2} = -4.5 \). So two points are \( (0, -3) \) and \( (-4.5, 0) \).

Step2: Analyze the second equation \( y = \frac{1}{3}x + 3 \)

For this line, when \( x = 0 \), \( y = 3 \). When \( y = 0 \), \( 0 = \frac{1}{3}x + 3 \), so \( \frac{1}{3}x = -3 \), \( x = -9 \). So two points are \( (0, 3) \) and \( (-9, 0) \).

Step3: Find the intersection point (graphically)

When we plot both lines, the intersection point is where the two lines cross. To find it algebraically (to confirm), set the two equations equal:
\( -\frac{2}{3}x - 3 = \frac{1}{3}x + 3 \)
Add \( \frac{2}{3}x \) to both sides: \( -3 = x + 3 \)
Subtract 3 from both sides: \( x = -6 \)
Substitute \( x = -6 \) into \( y = \frac{1}{3}x + 3 \): \( y = \frac{1}{3}(-6) + 3 = -2 + 3 = 1 \)
So the intersection point (solution) is \( (-6, 1) \).

Answer:

The solution to the system of equations is \( x = -6 \), \( y = 1 \) (or the point \( (-6, 1) \)).