Question

2 (a) graph both equations on the same graph: equation #1: $y = -x + 1$ equation #2: $-x + 3y = -9$

Explanation:

Step1: Analyze Equation #1: \( y = -x + 1 \)

This is in slope - intercept form (\( y=mx + b \)), where the slope \( m=-1 \) and the y - intercept \( b = 1 \).

• To find two points:
• When \( x = 0 \), \( y=-0 + 1=1 \), so the point is \( (0,1) \).
• When \( x = 1 \), \( y=-1 + 1=0 \), so the point is \( (1,0) \).

Step2: Analyze Equation #2: \( -x + 3y=-9 \)

First, rewrite it in slope - intercept form (\( y=mx + b \)):

• Add \( x \) to both sides: \( 3y=x - 9 \).
• Divide both sides by 3: \( y=\frac{1}{3}x-3 \).
• The slope \( m = \frac{1}{3} \) and the y - intercept \( b=-3 \).
• To find two points:
• When \( x = 0 \), \( y=\frac{1}{3}(0)-3=-3 \), so the point is \( (0, - 3) \).
• When \( x = 3 \), \( y=\frac{1}{3}(3)-3=1 - 3=-2 \), so the point is \( (3,-2) \).

Step3: Graph the lines

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

(Note: Since the question is to graph the equations, the final answer is the graph with the two lines plotted as described above. If we were to find the intersection point (for solving the system), we can set \( -x + 1=\frac{1}{3}x-3 \).

• Add \( x \) to both sides: \( 1=\frac{1}{3}x+x - 3 \).
• Combine like terms: \( 1=\frac{1 + 3}{3}x-3=\frac{4}{3}x-3 \).
• Add 3 to both sides: \( 4=\frac{4}{3}x \).
• Multiply both sides by \( \frac{3}{4} \): \( x = 3 \).
• Substitute \( x = 3 \) into \( y=-x + 1 \), \( y=-3 + 1=-2 \). So the intersection point is \( (3,-2) \). But the main task here is graphing.)

Answer:

The graph consists of two lines: one with slope - 1 and y - intercept 1 (passing through (0,1) and (1,0)) and another with slope \( \frac{1}{3} \) and y - intercept - 3 (passing through (0, - 3) and (3,-2)). The intersection point of the two lines is \( (3,-2) \).