Question

1. $y = -2x + 2$

$y = -2x - 2$

1. $y = \frac{1}{3}x - 3$

$y = -x + 1$

Explanation:

For Problem 6:

Step1: Identify slope-intercept form

Both equations use $y=mx+b$, where $m$=slope, $b$=y-intercept.
For $y=-2x+2$: $m=-2$, $b=2$
For $y=-2x-2$: $m=-2$, $b=-2$

Step2: Plot y-intercepts

For $y=-2x+2$: mark $(0, 2)$
For $y=-2x-2$: mark $(0, -2)$

Step3: Use slope to find 2nd points

Slope $-2=\frac{-2}{1}$: from $(0,2)$, move 1 right, 2 down to $(1,0)$; from $(0,-2)$, move 1 right, 2 down to $(1,-4)$

Step4: Draw lines

Connect points for each equation; lines are parallel.

Step1: Identify slope-intercept form

For $y=\frac{1}{3}x-3$: $m=\frac{1}{3}$, $b=-3$
For $y=-x+1$: $m=-1$, $b=1$

Step2: Plot y-intercepts

For $y=\frac{1}{3}x-3$: mark $(0, -3)$
For $y=-x+1$: mark $(0, 1)$

Step3: Use slope to find 2nd points

For $y=\frac{1}{3}x-3$: slope $\frac{1}{3}$ → move 3 right, 1 up to $(3,-2)$
For $y=-x+1$: slope $-1=\frac{-1}{1}$ → move 1 right, 1 down to $(1,0)$

Step4: Find intersection (solve system)

Set $\frac{1}{3}x-3 = -x+1$
$\frac{1}{3}x + x = 1 + 3$
$\frac{4}{3}x = 4$
$x=3$, substitute to get $y=-3+1=-2$
Intersection at $(3,-2)$

Step5: Draw lines

Connect points for each equation, crossing at $(3,-2)$.

Answer:

• Line $y=-2x+2$ passes through $(0,2)$ and $(1,0)$, extending across the grid.
• Line $y=-2x-2$ passes through $(0,-2)$ and $(1,-4)$, extending across the grid.
• The two lines are parallel (no intersection).

---

For Problem 8: