APPLYING THE MATHEMATICS 13. Al budgets \( \$ 100 \) for lunches during a month, and he estimates that each lunch costs \( \$ 5 \). a. After he has purchased \( x \) lunches, how much money does he have left? b. Let \( y \) be the amount of money he has left after he has purchased \( x \) lunches. Graph the relationship between \( y \) and \( x \). c. Use your graph to determine how many luches he can purchase and have at least \( \$ 20 \) left for a nice dinner. d. What inequality can be solved to give the answer to Part c?

Let's break down the problem step by step. ### a. Amount of Money Left After Purchasing \( x \) Lunches Al budgets \( \$ 100 \) for lunches, and each lunch costs \( \$ 5 \). After purchasing \( x \) lunches, the total cost for the lunches is \( 5x \). Therefore, the amount of money he has left, \( y \), can be expressed as: \[ y = 100 - 5x \] ### b. Graphing the Relationship Between \( y \) and \( x \) To graph the relationship between \( y \) and \( x \), we can plot the equation \( y = 100 - 5x \). 1. **Find the intercepts**: - **y-intercept**: Set \( x = 0 \): \[ y = 100 - 5(0) = 100 \] So, the y-intercept is \( (0, 100) \). - **x-intercept**: Set \( y = 0 \): \[ 0 = 100 - 5x \implies 5x = 100 \implies x = 20 \] So, the x-intercept is \( (20, 0) \). 2. **Plot the points**: Plot the points \( (0, 100) \) and \( (20, 0) \) on a graph. 3. **Draw the line**: Connect the points with a straight line. The line will slope downwards from left to right. ### c. Determining How Many Lunches He Can Purchase and Have at Least \( \$ 20 \) Left Al wants to have at least \( \$ 20 \) left after purchasing lunches. This means: \[ y \geq 20 \] Substituting \( y \) from the equation we derived: \[ 100 - 5x \geq 20 \] Now, solve for \( x \): \[ 100 - 20 \geq 5x \] \[ 80 \geq 5x \] \[ \frac{80}{5} \geq x \] \[ 16 \geq x \] This means Al can purchase a maximum of \( 16 \) lunches and still have at least \( \$ 20 \) left. ### d. Inequality to Solve for Part c The inequality that can be solved to find the maximum number of lunches Al can purchase while having at least \( \$ 20 \) left is: \[ 100 - 5x \geq 20 \] This inequality can be rearranged and solved as shown in part c.