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1) $y > \\frac{1}{3}x + 1$ $y < \\frac{4}{3}x - 2$ 12) $y \\geq \\frac{…

Question

  1. $y > \frac{1}{3}x + 1$

$y < \frac{4}{3}x - 2$

  1. $y \geq \frac{5}{2}x - 3$

$y \leq \frac{1}{2}x + 1$

  1. $y < -4x - 3$

$y \geq 2x + 3$

  1. $y > -3$

$y \leq -\frac{4}{3}x + 1$

Explanation:

Problem 11: $y>\frac{1}{3}x+1$ and $y<\frac{4}{3}x-2$

Step1: Graph boundary lines

  • For $y=\frac{1}{3}x+1$: slope $m=\frac{1}{3}$, y-intercept $(0,1)$. Draw dashed line (inequality is $>$).
  • For $y=\frac{4}{3}x-2$: slope $m=\frac{4}{3}$, y-intercept $(0,-2)$. Draw dashed line (inequality is $<$).

Step2: Shade solution regions

  • Shade above $y=\frac{1}{3}x+1$ (satisfies $y>$).
  • Shade below $y=\frac{4}{3}x-2$ (satisfies $y<$).
  • Overlapping shaded area is the solution.

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Problem 12: $y\geq\frac{5}{2}x-3$ and $y\leq-\frac{1}{2}x+1$

Step1: Graph boundary lines

  • For $y=\frac{5}{2}x-3$: slope $m=\frac{5}{2}$, y-intercept $(0,-3)$. Draw solid line (inequality is $\geq$).
  • For $y=-\frac{1}{2}x+1$: slope $m=-\frac{1}{2}$, y-intercept $(0,1)$. Draw solid line (inequality is $\leq$).

Step2: Shade solution regions

  • Shade above $y=\frac{5}{2}x-3$ (satisfies $y\geq$).
  • Shade below $y=-\frac{1}{2}x+1$ (satisfies $y\leq$).
  • Overlapping shaded area is the solution.

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Problem 13: $y<-4x-3$ and $y\geq2x+3$

Step1: Graph boundary lines

  • For $y=-4x-3$: slope $m=-4$, y-intercept $(0,-3)$. Draw dashed line (inequality is $<$).
  • For $y=2x+3$: slope $m=2$, y-intercept $(0,3)$. Draw solid line (inequality is $\geq$).

Step2: Shade solution regions

  • Shade below $y=-4x-3$ (satisfies $y<$).
  • Shade above $y=2x+3$ (satisfies $y\geq$).
  • Overlapping shaded area is the solution.

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Problem 14: $y>-3$ and $y\leq-\frac{4}{3}x+1$

Step1: Graph boundary lines

  • For $y=-3$: horizontal line through $(0,-3)$. Draw dashed line (inequality is $>$).
  • For $y=-\frac{4}{3}x+1$: slope $m=-\frac{4}{3}$, y-intercept $(0,1)$. Draw solid line (inequality is $\leq$).

Step2: Shade solution regions

  • Shade above $y=-3$ (satisfies $y>$).
  • Shade below $y=-\frac{4}{3}x+1$ (satisfies $y\leq$).
  • Overlapping shaded area is the solution.

Answer:

  1. For system 11: Dashed lines $y=\frac{1}{3}x+1$ and $y=\frac{4}{3}x-2$; solution is the region above the first line and below the second line.
  2. For system 12: Solid lines $y=\frac{5}{2}x-3$ and $y=-\frac{1}{2}x+1$; solution is the region above the first line and below the second line.
  3. For system 13: Dashed line $y=-4x-3$, solid line $y=2x+3$; solution is the region below the first line and above the second line.
  4. For system 14: Dashed horizontal line $y=-3$, solid line $y=-\frac{4}{3}x+1$; solution is the region above the horizontal line and below the second line.