QUESTION IMAGE
Question
question 14 (essay worth 10 points)
(graphing systems of inequalities hc)
the following system of inequalities is given
$x + 2y \geq 16$
$x + y < 20$
part a: explain how to graph the solution for the system of inequalities. be sure to include the type of line and the direction of the shading. (6 points)
part b: determine if the ordered pair $(-10, 20)$ is in the solution set. explain why or why not. (2 points)
part c: determine if the ordered pair $(20, 10)$ is in the solution set. explain why or why not. (2 points)
Part A
Step1: Rewrite inequalities to slope-intercept
For $x+2y\geq16$:
$2y\geq -x+16$
$y\geq \frac{-1}{2}x + 8$
For $x+y<20$:
$y < -x + 20$
Step2: Graph boundary lines
- For $y\geq \frac{-1}{2}x + 8$: Use a solid line (since inequality is $\geq$, includes boundary). Plot using intercepts: when $x=0$, $y=8$; when $y=0$, $x=16$.
- For $y < -x + 20$: Use a dashed line (since inequality is $<$, excludes boundary). Plot using intercepts: when $x=0$, $y=20$; when $y=0$, $x=20$.
Step3: Determine shading direction
- For $y\geq \frac{-1}{2}x + 8$: Shade above the solid line (test point $(0,0)$: $0\geq 8$ is false, so shade opposite side).
- For $y < -x + 20$: Shade below the dashed line (test point $(0,0)$: $0 < 20$ is true, so shade this side).
- The solution is the overlapping shaded region.
Part B
Step1: Substitute $(-10,20)$ into inequalities
First inequality: $x+2y\geq16$
$30\geq16$ is true.
Second inequality: $x+y<20$
$10<20$ is true.
Part C
Step1: Substitute $(20,10)$ into inequalities
First inequality: $x+2y\geq16$
$40\geq16$ is true.
Second inequality: $x+y<20$
$30<20$ is false.
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Part A
- Rewrite both inequalities to slope-intercept form: $y\geq \frac{-1}{2}x + 8$ and $y < -x + 20$.
- Graph a solid line for $y\geq \frac{-1}{2}x + 8$ (includes boundary) and a dashed line for $y < -x + 20$ (excludes boundary).
- Shade above the solid line and below the dashed line; the overlapping shaded area is the solution.
Part B
The ordered pair $(-10, 20)$ is in the solution set. It satisfies both inequalities: $-10+2(20)=30\geq16$ and $-10+20=10<20$.
Part C
The ordered pair $(20, 10)$ is NOT in the solution set. While it satisfies $20+2(10)=40\geq16$, it fails $20+10=30<20$, which is false.