Question

ear inequalities.
is that are solutions.
irs that are not solutions.

1. \

$$\begin{cases} y \\leq \\dfrac{1}{2}x + 1 \\\\ x + y < 3 \\end{cases}$$

Explanation:

Step1: Graph first inequality boundary

The boundary of $y \leq \frac{1}{2}x + 1$ is the line $y = \frac{1}{2}x + 1$. It has a y-intercept at $(0,1)$ and slope $\frac{1}{2}$. Since the inequality is $\leq$, the line is solid, and we shade below it.

Step2: Graph second inequality boundary

The boundary of $x + y < 3$ is the line $x + y = 3$, or $y = -x + 3$. It has a y-intercept at $(0,3)$ and slope $-1$. Since the inequality is $<$, the line is dashed, and we shade below it.

Step3: Identify solution region

The solution region is the overlapping shaded area of both inequalities.

Step4: List solution points

Pick points in the overlap:

• $(0,0)$: $0 \leq \frac{1}{2}(0)+1$ (true, $0 \leq 1$) and $0+0 < 3$ (true, $0 < 3$)
• $(2,0)$: $0 \leq \frac{1}{2}(2)+1$ (true, $0 \leq 2$) and $2+0 < 3$ (true, $2 < 3$)

Step5: List non-solution points

Pick points not in the overlap:

• $(0,4)$: $4 \leq \frac{1}{2}(0)+1$ (false, $4 > 1$)
• $(4,2)$: $2 \leq \frac{1}{2}(4)+1$ (true, $2 \leq 3$) but $4+2 < 3$ (false, $6 > 3$)

Answer:

Solution Points:

$(0,0)$, $(2,0)$

Non-Solution Points:

$(0,4)$, $(4,2)$

Graph Description:

1. Solid line for $y = \frac{1}{2}x + 1$, shade below the line.
2. Dashed line for $y = -x + 3$, shade below the line.
3. The overlapping shaded area is the solution set of the system.