Question

solve the following system of equations graphically on the set of axes below.
$y = -\frac{2}{3}x - 3$
$x - 3y = -9$
plot two lines by clicking the graph.
click a line to delete it.
answer attempt 1 out of a
solution:

Explanation:

Step 1: Analyze the first equation \( y = -\frac{2}{3}x - 3 \)

This is in slope - intercept form (\(y=mx + b\)), where the slope \(m = -\frac{2}{3}\) and the y - intercept \(b=-3\). To graph this line, we can start at the y - intercept \((0,-3)\). Then, using the slope, from \((0, - 3)\), we can move down 2 units and right 3 units (or up 2 units and left 3 units) to find another point on the line.

Step 2: Analyze the second equation \(x - 3y=-9\)

We can rewrite this equation in slope - intercept form. Solve for \(y\):
\[

$$\begin{align*} x-3y&=-9\\ - 3y&=-x - 9\\ y&=\frac{1}{3}x + 3 \end{align*}$$

\]
Here, the slope \(m=\frac{1}{3}\) and the y - intercept \(b = 3\). To graph this line, we start at the y - intercept \((0,3)\). Then, using the slope, from \((0,3)\), we can move up 1 unit and right 3 units (or down 1 unit and left 3 units) to find another point on the line.

Step 3: Find the intersection point

When we graph both lines, we need to find the point where they intersect. Let's also solve the system algebraically to confirm (since we can use the intersection of the graphs as the solution). We have the system:
\[

$$\begin{cases} y=-\frac{2}{3}x - 3\\ y=\frac{1}{3}x+3 \end{cases}$$

\]
Set the two expressions for \(y\) equal to each other:
\[
-\frac{2}{3}x-3=\frac{1}{3}x + 3
\]
Add \(\frac{2}{3}x\) to both sides:
\[
-3=\frac{1}{3}x+\frac{2}{3}x + 3
\]
Simplify the right - hand side: \(\frac{1 + 2}{3}x=\ x\), so \(-3=x + 3\)
Subtract 3 from both sides: \(x=-6\)
Now substitute \(x = - 6\) into \(y=\frac{1}{3}x + 3\):
\[
y=\frac{1}{3}(-6)+3=-2 + 3=1
\]

Answer:

The solution of the system of equations (the point of intersection of the two lines) is \((-6,1)\)