Bashir is working two summer jobs, making \( \$ 17 \) per hour lifeguarding and making \( \$ 10 \) per hour washing cars. In a given week, he can work no more than 15 total hours and must earn at least \( \$ 200 \). If \( x \) represents the number of hours lifeguarding and \( y \) represents the number of hours washing cars, write and solve a system of inequalities graphically and determine one possible solution. Inequality \( 1: y \geq \) Inequality \( 2: y \geq \)

To model Bashir's work situation with inequalities, we need to consider the constraints based on his working hours and earnings. 1. **Total hours constraint**: Bashir can work no more than 15 total hours. This can be expressed as: \[ x + y \leq 15 \] 2. **Earnings constraint**: Bashir must earn at least $200. The earnings from lifeguarding and washing cars can be expressed as: \[ 17x + 10y \geq 200 \] Now, we can rewrite these inequalities: 1. From the total hours constraint: \[ y \leq 15 - x \] (This is the first inequality.) 2. From the earnings constraint: \[ 10y \geq 200 - 17x \implies y \geq \frac{200 - 17x}{10} \] (This is the second inequality.) Now we have the system of inequalities: 1. \( y \leq 15 - x \) 2. \( y \geq \frac{200 - 17x}{10} \) ### Graphing the Inequalities To graph these inequalities, we can find the intercepts and plot the lines. 1. **For \( y = 15 - x \)**: - When \( x = 0 \), \( y = 15 \) (point (0, 15)). - When \( y = 0 \), \( x = 15 \) (point (15, 0)). - This line will be a downward slope. 2. **For \( y = \frac{200 - 17x}{10} \)**: - When \( x = 0 \), \( y = 20 \) (point (0, 20)). - When \( y = 0 \), \( 200 - 17x = 0 \implies x = \frac{200}{17} \approx 11.76 \) (point (11.76, 0)). - This line will be an upward slope. ### Finding the Intersection Points To find the feasible region, we need to find the intersection of the two lines: Set \( 15 - x = \frac{200 - 17x}{10} \): \[ 10(15 - x) = 200 - 17x \] \[ 150 - 10x = 200 - 17x \] \[ 7x = 50 \implies x = \frac{50}{7} \approx 7.14 \] Now substitute \( x \) back to find \( y \): \[ y = 15 - \frac{50}{7} = \frac{105 - 50}{7} = \frac{55}{7} \approx 7.86 \] So the intersection point is approximately \( \left( \frac{50}{7}, \frac{55}{7} \right) \). ### Possible Solution Now we can choose a point in the feasible region. One simple solution is to check the point where \( x = 10 \) and \( y = 5 \): - Check the total hours: \( 10 + 5 = 15 \) (satisfies \( x + y \leq 15 \)). - Check the earnings: \( 17(10) + 10(5) = 170 + 50 = 220 \) (satisfies \( 17x + 10y \geq 200 \)). Thus, one possible solution is: \[ (x, y) = (10, 5) \] This means Bashir can work 10 hours as a lifeguard and 5 hours washing cars.