Question

1. $y \geq x - 3$

$y \geq -x - 1$

1. $x \leq -3$

$y < -|x| + 1$

1. $y > |x + 2|$

$2y < -x + 4$

Explanation:

4) System: $y \geq x - 3$, $y \geq -x - 1$

Step1: Graph $y=x-3$

Identify intercepts: when $x=0$, $y=-3$; when $y=0$, $x=3$. Draw a solid line (due to $\geq$). Shade above the line.

Step2: Graph $y=-x-1$

Identify intercepts: when $x=0$, $y=-1$; when $y=0$, $x=-1$. Draw a solid line (due to $\geq$). Shade above the line.

Step3: Find overlap region

The solution is the area shaded by both inequalities, above both lines. Their intersection is found by solving $x-3=-x-1$:
$2x=2 \implies x=1$, substitute to get $y=1-3=-2$. Intersection point: $(1, -2)$.

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5) System: $x \leq -3$, $y < -|x| + 1$

Step1: Graph $x=-3$

Draw a solid vertical line at $x=-3$ (due to $\leq$). Shade to the left of the line.

Step2: Graph $y=-|x|+1$

This is a downward-opening V with vertex at $(0,1)$. When $x=0$, $y=1$; when $y=0$, $x=\pm1$. Draw a dashed line (due to $<$). Shade below the line.

Step3: Find overlap region

The solution is the area shaded by both: left of $x=-3$ and below $y=-|x|+1$.

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6) System: $y > |x+2|$, $2y < -x + 4$ (rewrite as $y < \frac{-x+4}{2}$)

Step1: Graph $y=|x+2|$

This is a right-opening V with vertex at $(-2,0)$. Draw a dashed line (due to $>$). Shade above the line.

Step2: Graph $y=\frac{-x+4}{2}$

Identify intercepts: when $x=0$, $y=2$; when $y=0$, $x=4$. Draw a dashed line (due to $<$). Shade below the line.

Step3: Find overlap region

Find intersection by solving $|x+2|=\frac{-x+4}{2}$:
Case 1: $x+2 \geq 0$ ($x \geq -2$):
$x+2=\frac{-x+4}{2} \implies 2x+4=-x+4 \implies 3x=0 \implies x=0$, $y=2$. Point $(0,2)$.
Case 2: $x+2 < 0$ ($x < -2$):
$-x-2=\frac{-x+4}{2} \implies -2x-4=-x+4 \implies -x=8 \implies x=-8$, $y=|-8+2|=6$. Point $(-8,6)$.
The solution is the area above $y=|x+2|$ and below $y=\frac{-x+4}{2}$, between $x=-8$ and $x=0$.

Answer:

1. For $y \geq x - 3$ and $y \geq -x - 1$:

• Solid lines for $y=x-3$ (intercepts $(0,-3),(3,0)$) and $y=-x-1$ (intercepts $(0,-1),(-1,0)$), shaded region above both lines, intersecting at $(1,-2)$.

1. For $x \leq -3$ and $y < -|x| + 1$:

• Solid vertical line $x=-3$ (shaded left) and dashed V-shaped line $y=-|x|+1$ (vertex $(0,1)$, intercepts $(\pm1,0)$), shaded region left of $x=-3$ and below the V.

1. For $y > |x+2|$ and $2y < -x + 4$:

• Dashed V-shaped line $y=|x+2|$ (vertex $(-2,0)$) shaded above, dashed line $y=\frac{-x+4}{2}$ (intercepts $(0,2),(4,0)$) shaded below, overlapping region between intersection points $(-8,6)$ and $(0,2)$.