Question

(Look at picture) Ty for the help

Answer 1

Inequalities:

• x + y ≤ 35
• 15x + 10y ≥ 400

`\begin{array}\cline{1-3} \vphantom{\frac12} & \sf Viable & \sf Nonviable\\\cline{1-3} \vphantom{\frac12} (10,25)& \checkmark & \\\cline{1-3} \vphantom{\frac12} (10,20)&&\checkmark \\\cline{1-3} \vphantom{\frac12} (20,12)&\checkmark& \\\cline{1-3} \vphantom{\frac12} (35,0)&\checkmark& \\\cline{1-3} \vphantom{\frac12} (20,20)&&\checkmark \\\cline{1-3}\end{array}`

Explanation:

Part A

Definition of variables:

• Let x = the number of hours cutting grass.
• Let y = the number of hours tutoring.

Given information:

• Cutting grass = $15 per hour
• Tutoring reading = $10 per hour
• Save at least $400
• Work no more than 35 hours

The inequality that represents Barrett working no more than 35 hours is:

• x + y ≤ 35

The inequality that represents Barrett earning at least $400 is:

• 15x + 10y ≥ 400

Part B

To determine if each point is a viable or nonviable solution, substitute each point into the two inequalities. If both inequalities are true, the solution is viable.

`\begin{aligned}(10,25) \implies 10+25 &\leq 35\\35& \leq 35 \leftarrow \sf true\\\\ \implies 15(10)+10(25)&\geq400\\400&\geq 400 \leftarrow \sf true\end{aligned}`

Therefore, point (10, 25) is a viable solution.

`\begin{aligned}(10,20) \implies 10+20 &\leq 35\\30& \leq 35 \leftarrow \sf true\\\\ \implies 15(10)+10(20)&\geq400\\350&\geq 400 \leftarrow \sf false\end{aligned}`

Therefore, point (10, 20) is not a viable solution.

`\begin{aligned}(20,12) \implies 20+12 &\leq 35\\32& \leq 35 \leftarrow \sf true\\\\ \implies 15(20)+10(12)&\geq400\\420&\geq 400 \leftarrow \sf true\end{aligned}`

Therefore, point (20, 12) is a viable solution.

`\begin{aligned}(35,0) \implies 35+0 &\leq 35\\32& \leq 35 \leftarrow \sf true\\\\ \implies 15(35)+10(0)&\geq400\\525&\geq 400 \leftarrow \sf true\end{aligned}`

Therefore, point (35,0) is a viable solution.

`\begin{aligned}(20,20) \implies 20+20 &\leq 35\\40& \leq 35 \leftarrow \sf false\\\\ \implies 15(20)+10(20)&\geq400\\500&\geq 400 \leftarrow \sf true\end{aligned}`

Therefore, point (20, 20) is not a viable solution.

`\begin{array}c\cline{1-3} \vphantom{\frac12} & \sf Viable & \sf Nonviable\\\cline{1-3} \vphantom{\frac12} (10,25)& \checkmark & \\\cline{1-3} \vphantom{\frac12} (10,20)&&\checkmark \\\cline{1-3} \vphantom{\frac12} (20,12)&\checkmark& \\\cline{1-3} \vphantom{\frac12} (35,0)&\checkmark& \\\cline{1-3} \vphantom{\frac12} (20,20)&&\checkmark \\\cline{1-3}\end{array}`