Question

1. you can work at most 20 hours next week. you need to earn at least $92 to cover your weekly expenses. your dog-walking job pays $7.50 per hour and your job as a car wash attendant pays $6.00 per hour. write and graph a system of linear inequalities to model the situation. what are three possible solutions to this answer? possible solutions to this answer? x-hours of dog

Explanation:

Step1: Define Variables

Let \( x \) be the number of hours spent dog - walking and \( y \) be the number of hours spent as a car wash attendant.

Step2: Time Constraint Inequality

The total number of hours worked (\( x + y \)) can be at most 20 hours. So, the inequality for the time constraint is \( x + y\leqslant20 \), which can be rewritten as \( y\leqslant - x + 20 \).

Step3: Earnings Constraint Inequality

The dog - walking job pays \( \$7.50 \) per hour and the car wash job pays \( \$6.00 \) per hour. The total earnings (\( 7.5x+6y \)) need to be at least \( \$92 \). So, the inequality for the earnings constraint is \( 7.5x + 6y\geqslant92 \), which can be rewritten as \( y\geqslant-\frac{7.5}{6}x+\frac{92}{6}\), or simplifying \( y\geqslant - 1.25x+\frac{46}{3}\approx - 1.25x + 15.33\). Also, since the number of hours cannot be negative, we have \( x\geqslant0 \) and \( y\geqslant0 \).

Step4: Find Possible Solutions

We can find solutions by choosing values of \( x \) (number of dog - walking hours) and \( y \) (number of car - wash hours) that satisfy all the inequalities: \( x\geqslant0 \), \( y\geqslant0 \), \( x + y\leqslant20 \), and \( 7.5x+6y\geqslant92 \).

• Solution 1: Let's start with \( x = 8 \).

Substitute \( x = 8 \) into the earnings inequality: \( 7.5\times8+6y\geqslant92\)
\( 60 + 6y\geqslant92\)
\( 6y\geqslant92 - 60=32\)
\( y\geqslant\frac{32}{6}\approx5.33\)
Substitute \( x = 8 \) into the time inequality: \( 8 + y\leqslant20\)
\( y\leqslant12\)
Since \( y\) is a non - negative integer (number of hours), we can take \( y = 6 \). Check: \( 7.5\times8+6\times6=60 + 36 = 96\geqslant92\) and \( 8 + 6=14\leqslant20\).

• Solution 2: Let \( x = 10 \)

Substitute into earnings inequality: \( 7.5\times10+6y\geqslant92\)
\( 75+6y\geqslant92\)
\( 6y\geqslant17\)
\( y\geqslant\frac{17}{6}\approx2.83\)
Substitute into time inequality: \( 10 + y\leqslant20\)
\( y\leqslant10\)
Take \( y = 3 \). Check: \( 7.5\times10+6\times3=75 + 18=93\geqslant92\) and \( 10 + 3 = 13\leqslant20\).

• Solution 3: Let \( x = 4 \)

Substitute into earnings inequality: \( 7.5\times4+6y\geqslant92\)
\( 30+6y\geqslant92\)
\( 6y\geqslant62\)
\( y\geqslant\frac{62}{6}\approx10.33\)
Substitute into time inequality: \( 4 + y\leqslant20\)
\( y\leqslant16\)
Take \( y = 11 \). Check: \( 7.5\times4+6\times11=30 + 66 = 96\geqslant92\) and \( 4+11 = 15\leqslant20\).

Answer:

The system of linear inequalities is:
\(

$$\begin{cases}x\geqslant0\\y\geqslant0\\x + y\leqslant20\\7.5x+6y\geqslant92\end{cases}$$

\)

Three possible solutions are:

1. \(x = 8\), \(y = 6\) (8 hours of dog - walking and 6 hours of car - wash)
2. \(x = 10\), \(y = 3\) (10 hours of dog - walking and 3 hours of car - wash)
3. \(x = 4\), \(y = 11\) (4 hours of dog - walking and 11 hours of car - wash)