Problem Analysis
The problem involves solving a system of inequalities to determine how Bobbi can earn at least \$150 while working at most 20 hours next week. She has two jobs:
1.
Dog-walking job: Pays \$6 per hour.
2.
Car washing job: Pays \$10 per hour.
We are given the following constraints:
1.
Total hours worked: \( x + y \leq 20 \)
2.
Minimum earnings: \( 6x + 10y \geq 150 \)
Where:
- \( x \): Hours spent dog-walking
- \( y \): Hours spent washing cars
Step-by-Step Solution
#### 1. Graph the First Inequality: \( x + y \leq 20 \)
This inequality represents the total number of hours Bobbi can work. To graph it:
- Rewrite the inequality as an equation: \( x + y = 20 \).
- Find two points on this line:
- When \( x = 0 \): \( y = 20 \). Point: \( (0, 20) \)
- When \( y = 0 \): \( x = 20 \). Point: \( (20, 0) \)
- Plot these points and draw a line through them. Since the inequality is \( \leq \), shade the region below the line (including the line).
#### 2. Graph the Second Inequality: \( 6x + 10y \geq 150 \)
This inequality represents the minimum earnings Bobbi needs. To graph it:
- Rewrite the inequality as an equation: \( 6x + 10y = 150 \).
- Simplify by dividing through by 2: \( 3x + 5y = 75 \).
- Find two points on this line:
- When \( x = 0 \): \( 5y = 75 \Rightarrow y = 15 \). Point: \( (0, 15) \)
- When \( y = 0 \): \( 3x = 75 \Rightarrow x = 25 \). Point: \( (25, 0) \)
- Plot these points and draw a line through them. Since the inequality is \( \geq \), shade the region above the line (including the line).
#### 3. Identify the Feasible Region
The feasible region is the area where both inequalities are satisfied simultaneously. This is the intersection of the shaded regions from the two inequalities.
#### 4. Verify a Solution
The problem suggests one possible solution: Bobbi works 6 hours dog-walking (\( x = 6 \)) and 12 hours washing cars (\( y = 12 \)). Let's verify this solution:
1.
Check the total hours constraint:
\[
x + y = 6 + 12 = 18 \leq 20
\]
This satisfies \( x + y \leq 20 \).
2.
Check the minimum earnings constraint:
\[
6x + 10y = 6(6) + 10(12) = 36 + 120 = 156 \geq 150
\]
This satisfies \( 6x + 10y \geq 150 \).
Since both constraints are satisfied, \( (x, y) = (6, 12) \) is a valid solution.
Final Answer
\[
\boxed{(6, 12)}
\]
Bobbi can work 6 hours walking dogs and 12 hours washing cars to meet her requirements.
Parent Tip: Review the logic above to help your child master the concept of system of linear inequalities word problems worksheet.
