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# Help with this one CPhill ​

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Help with this one CPhill

Nov 21, 2018

### 5+0 Answers

#1
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I do not think Chris is around at this time, so I will try to help you. CPhill can still take a look at your question for further help.

Note: A solid line ( _____) means that the variable is $$\ge$$ or $$\le$$ to the value. A dotted line (---------) means that the variable is $$>$$  or $$<$$ to the value.

A) We have the inequality $$y\ge5$$, which states that no matter what x is, y is at least 5. We can draw a solid line parallel to the x-axis and intersecting the y-axis at y = 5, and anything above the y = 5 line is shaded.

We also know that $$y\le \dfrac{2}{3}x + 3$$. We can treat this as a line with the equation $$y = \dfrac{2}{3}x + 3$$. We can determine the line by identifying two points. If we try x = 0, we have y = 3, and if we try x = 3, we have $$y = \dfrac{2}{3}\cdot3 + 3$$, which simplifies into y = 5. Our two coordinate points are (0, 3) and (3, 5). We can draw a solid line connecting the two points, and that is the line $$y = \dfrac{2}{3}x + 3$$. We are not done yet though, we still need to shade the area above or below $$y = \dfrac{2}{3}x + 3$$. The original inequality is $$y \le \dfrac{2}{3}x + 3$$, and we can try some points to find the right boundary. Let's try the point (0, 0). Our inequality is now $$0 \le \dfrac{2}{3}\cdot0 + 3$$, which simplifies into $$0 \le 3$$. This statement is true, so we shade in the bottom side with (0, 0).

B) Solutions to both inequalities would have to lie in both shaded boundaries. Looking at the graph (below in next answer), we can clearly see that (6, 6) is in both shaded boundaries. One solution is the point (6, 6). (There are many other solutions, such as (3, 5) and (8, 8).)

Hope this helps,

- PartialMathematician

Nov 21, 2018
edited by PartialMathematician  Nov 21, 2018
edited by PartialMathematician  Nov 21, 2018
#2
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Here is the link to the graph:

https://www.desmos.com/calculator/tclz4zr88j

Hope this helps,

- PartialMathematician

Nov 21, 2018
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CPhill, please check my answers to make sure I gave "NotSoSmart" the correct answers.

Thanks,

-PM

PartialMathematician  Nov 21, 2018
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You can take PM's answer to the bank, NSS

Excellent presentation, PM.......!!!!!

Nov 21, 2018
#5
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Thanks, CPhill!

PartialMathematician  Nov 21, 2018