Question

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

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 $y = 0$, $0=-x + 1\Rightarrow x = 1$. So the point is $(1,0)$.

Step2: Rewrite Equation #2 ($-x + 3y=-9$) in slope - intercept form

Solve for $y$:
\[

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

\]
Here, the slope $m = \frac{1}{3}$ and the y - intercept $b=-3$. To find two points:

• When $x = 0$, $y=\frac{1}{3}(0)-3=-3$. So the point is $(0,-3)$.
• When $y = 0$, $0=\frac{1}{3}x-3\Rightarrow\frac{1}{3}x = 3\Rightarrow x = 9$. So the point is $(9,0)$.

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 $(9,0)$ and draw a straight line through them.

(Note: Since the problem asks to graph the equations, the final answer is the graphical representation of the two lines. The key points for graphing are: Line 1: passes through $(0,1)$ and $(1,0)$ with slope - 1; Line 2: passes through $(0,-3)$ and $(9,0)$ with slope $\frac{1}{3}$)

Answer:

To graph the equations:

• Equation \(y = -x + 1\): Plot \((0,1)\) (y - intercept) and \((1,0)\) (x - intercept), then draw a line through these points (slope \(=-1\)).
• Equation \(-x + 3y=-9\) (or \(y=\frac{1}{3}x - 3\)): Plot \((0,-3)\) (y - intercept) and \((9,0)\) (x - intercept), then draw a line through these points (slope \(=\frac{1}{3}\)).