Question

solve each system of inequalities by graphing.

1. \\(\

$$\begin{cases}x \\leq -2 \\\\ x + y > 1\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y \\leq -3x + 1 \\\\ 3x + 2y > -12\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y \\geq x^2 - 4 \\\\ y < -x^2 - x + 2\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y < 4 \\\\ y \\leq |x - 1| - 3\\end{cases}$$

\\)

Explanation:

For Problem 11 (already partially solved, confirming the solution):

Step1: Graph $x \leq -2$

Draw a solid vertical line at $x=-2$, shade left of the line.

Step2: Rewrite $x+y>1$ to slope-intercept

$y > -x + 1$

Step3: Graph $y > -x + 1$

Draw a dashed line with slope $m=-1$, y-intercept $(0,1)$, shade above the line.

Step4: Identify overlapping region

The green shaded area where both regions intersect.

Step1: Rewrite $y \leq -3x + 1$

Graph solid line: $m=-3$, y-intercept $(0,1)$, shade below.

Step2: Rewrite $3x+2y>-12$

$2y > -3x -12 \implies y > -\frac{3}{2}x -6$

Step3: Graph $y > -\frac{3}{2}x -6$

Draw dashed line: $m=-\frac{3}{2}$, y-intercept $(0,-6)$, shade above.

Step4: Find intersection of lines

Solve

$$\begin{cases} y=-3x+1 \\ y=-\frac{3}{2}x-6 \end{cases}$$

$-3x+1 = -\frac{3}{2}x -6 \implies -\frac{3}{2}x = -7 \implies x=\frac{14}{3} \approx 4.67$, $y=-3(\frac{14}{3})+1=-13$
Intersection point: $(\frac{14}{3}, -13)$

Step5: Identify overlapping region

Shade the area that is below $y=-3x+1$ and above $y=-\frac{3}{2}x-6$.

Step1: Graph $y \geq x^2 -4$

Solid parabola: vertex $(0,-4)$, opens upward, shade above.

Step2: Graph $y < -x^2 -x +2$

Find vertex: $x=-\frac{b}{2a}=-\frac{-1}{2(-1)}=-\frac{1}{2}$, $y=-(-\frac{1}{2})^2 -(-\frac{1}{2})+2=\frac{9}{4}$
Dashed parabola: vertex $(-\frac{1}{2}, \frac{9}{4})$, opens downward, shade below.

Step3: Find intersection points

Solve $x^2 -4 = -x^2 -x +2$
$2x^2 +x -6=0$
Use quadratic formula: $x=\frac{-1\pm\sqrt{1+48}}{4}=\frac{-1\pm7}{4}$
$x=\frac{6}{4}=\frac{3}{2}$ or $x=\frac{-8}{4}=-2$
Corresponding $y$-values:
For $x=\frac{3}{2}$: $y=(\frac{3}{2})^2 -4=\frac{9}{4}-4=-\frac{7}{4}$
For $x=-2$: $y=(-2)^2 -4=0$
Intersection points: $(-2,0)$ and $(\frac{3}{2}, -\frac{7}{4})$

Step4: Identify overlapping region

Shade the area between the two parabolas, bounded by $(-2,0)$ and $(\frac{3}{2}, -\frac{7}{4})$.

Answer:

The solution is the set of all points $(x,y)$ where $x \leq -2$ and $y > -x + 1$, represented by the green overlapping region on the graph.

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For Problem 12: