Question

9 a system of equations is shown. \\(\

$$\begin{cases}x + 3y = -3 \\\\ 3x + 2y = 12\\end{cases}$$

\\) part a graph the system of equations.

Explanation:

Step1: Rewrite first equation in slope - intercept form

The first equation is \(x + 3y=-3\). We solve for \(y\):
Subtract \(x\) from both sides: \(3y=-x - 3\)
Divide both sides by 3: \(y =-\frac{1}{3}x-1\)
The slope \(m_1 =-\frac{1}{3}\) and the \(y\) - intercept \(b_1=-1\). To find two points on this line, when \(x = 0\), \(y=-1\) (the \(y\) - intercept). When \(y = 0\), \(0=-\frac{1}{3}x-1\), then \(\frac{1}{3}x=-1\), so \(x=-3\). So two points on the first line are \((0, - 1)\) and \((-3,0)\).

Step2: Rewrite second equation in slope - intercept form

The second equation is \(3x + 2y = 12\). We solve for \(y\):
Subtract \(3x\) from both sides: \(2y=-3x + 12\)
Divide both sides by 2: \(y=-\frac{3}{2}x + 6\)
The slope \(m_2=-\frac{3}{2}\) and the \(y\) - intercept \(b_2 = 6\). To find two points on this line, when \(x = 0\), \(y = 6\) (the \(y\) - intercept). When \(y=0\), \(0=-\frac{3}{2}x + 6\), then \(\frac{3}{2}x=6\), so \(x = 4\). So two points on the second line are \((0,6)\) and \((4,0)\).

Step3: Graph the lines

• For the line \(y =-\frac{1}{3}x-1\), plot the points \((0,-1)\) and \((-3,0)\) and draw a straight line through them.
• For the line \(y=-\frac{3}{2}x + 6\), plot the points \((0,6)\) and \((4,0)\) and draw a straight line through them.

(Note: Since this is a text - based response, we can't draw the graph here, but the steps above describe how to find the points to plot the two lines on the given coordinate plane.)

Answer:

To graph the system:

1. For \(x + 3y=-3\) (or \(y =-\frac{1}{3}x-1\)): Plot \((0,-1)\) and \((-3,0)\) and draw a line through them.
2. For \(3x + 2y = 12\) (or \(y=-\frac{3}{2}x + 6\)): Plot \((0,6)\) and \((4,0)\) and draw a line through them.