Question

question
abdoulaye has x quarters and y dimes. he has at least 16 coins worth no more than $3 combined. solve this system of inequalities graphically and determine one possible solution.
inequality 1: y ≥
inequality 2: y ≥
graph with y-axis and x-axis labeled, grid lines

Explanation:

Step1: Define the inequalities

First, the number of coins: \( x + y \geq 16 \), so \( y \geq -x + 16 \).
Second, the value of coins: A quarter is $0.25, a dime is $0.10. Total value \( 0.25x + 0.10y \leq 3 \). Multiply by 100: \( 25x + 10y \leq 300 \), simplify: \( 5x + 2y \leq 60 \), then \( y \leq -\frac{5}{2}x + 30 \).

Step2: Graph the inequalities

• For \( y \geq -x + 16 \): Plot the line \( y = -x + 16 \) (slope -1, y-intercept 16), shade above the line (since \( y \geq \)).
• For \( y \leq -\frac{5}{2}x + 30 \): Plot the line \( y = -\frac{5}{2}x + 30 \) (slope -2.5, y-intercept 30), shade below the line (since \( y \leq \)).

Step3: Find a possible solution

Find integer \( x, y \) in the overlapping shaded region. Let's pick \( x = 4 \). Then from \( y \geq -4 + 16 = 12 \) and \( y \leq -\frac{5}{2}(4) + 30 = -10 + 30 = 20 \). So \( y = 12 \) (or any between 12 and 20). Check: \( 4 + 12 = 16 \geq 16 \), value: \( 0.25(4) + 0.10(12) = 1 + 1.2 = 2.2 \leq 3 \). Valid.

Answer:

One possible solution is \( x = 4 \) quarters and \( y = 12 \) dimes (or other valid \( (x, y) \) in the feasible region). The inequalities are \( y \geq -x + 16 \) (Inequality 1) and \( y \leq -2.5x + 30 \) (Inequality 2).