Question

4 directions - solve the system by graphing. \\(\

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

\\)

Explanation:

Step1: Find intercepts for Line 1 ($y = \frac{1}{3}x - 3$)

• y-intercept: Set $x = 0$, then $y = \frac{1}{3}(0) - 3 = -3$. So point is $(0, -3)$.
• x-intercept: Set $y = 0$, then $0 = \frac{1}{3}x - 3$. Solve: $\frac{1}{3}x = 3 \implies x = 9$. So point is $(9, 0)$.

Step2: Find intercepts for Line 2 ($y = -x + 1$)

• y-intercept: Set $x = 0$, then $y = -0 + 1 = 1$. So point is $(0, 1)$.
• x-intercept: Set $y = 0$, then $0 = -x + 1 \implies x = 1$. So point is $(1, 0)$.

Step3: Graph the lines and find intersection

Plot the points for each line and draw the lines. The lines intersect where $x$ and $y$ satisfy both equations. Solve the system algebraically to confirm:
Set $\frac{1}{3}x - 3 = -x + 1$.
Add $x$ to both sides: $\frac{1}{3}x + x - 3 = 1 \implies \frac{4}{3}x - 3 = 1$.
Add 3 to both sides: $\frac{4}{3}x = 4$.
Multiply by $\frac{3}{4}$: $x = 3$.
Substitute $x = 3$ into $y = -x + 1$: $y = -3 + 1 = -2$. So intersection is $(3, -2)$.

Answer:

The solution to the system is $(3, -2)$ (the point where the two lines intersect when graphed).