Question

solve the system of inequalities by graphing.
$y \leq 1$
$y \geq -3x + 10$
select a line to change it between solid and dotted. select a region to shade it.

Explanation:

Step1: Identify line types

Both inequalities use $\leq$ and $\geq$, so both lines are solid (already correctly set in the graph: horizontal solid line for $y=1$, solid line for $y=-3x+10$).

Step2: Find intersection point

Set $y=1$ equal to $y=-3x+10$:
$$1 = -3x + 10$$
Solve for $x$:
$$3x = 10 - 1$$
$$x = \frac{9}{3} = 3$$
Intersection: $(3, 1)$

Step3: Shade for $y \leq 1$

Shade below the horizontal line $y=1$ (since $y$ is less than or equal to 1).

Step4: Shade for $y \geq -3x + 10$

Shade above the line $y=-3x+10$ (since $y$ is greater than or equal to the line's values).

Step5: Identify overlap region

The solution is the overlapping shaded area: the region that is both below $y=1$ and above $y=-3x+10$, bounded by the two lines and their intersection at $(3,1)$.

Answer:

1. Both lines remain solid (correct as shown).
2. Shade the region that is below the line $y=1$ and above the line $y=-3x+10$ (the small bounded area to the left of $x=3$, between the two lines, up to $y=1$ and down to the intersection point $(3,1)$).
3. The solution set is all points $(x,y)$ such that $x \leq 3$, $-3x+10 \leq y \leq 1$.