Question

eli and his children went into a restaurant where they sell drinks for $2 each and tacos for $3.50 each. eli has $45 to spend and must buy no less than 15 drinks and tacos altogether. write two possible solutions and relate them to the situation given. answer is a possible solution, meaning that eli can buy drinks and is a possible solution, meaning that eli can buy drinks and tacos.

Explanation:

Let \( d \) be the number of drinks and \( t \) be the number of tacos. The cost equation is \( 2d + 3.5t \leq 45 \), and the quantity constraint is \( d + t \geq 15 \).

Step 1: First Solution

Assume \( d = 10 \), then from \( d + t \geq 15 \), \( t \geq 5 \). Check cost: \( 2(10)+3.5t = 20 + 3.5t \leq 45 \Rightarrow 3.5t \leq 25 \Rightarrow t \leq \frac{25}{3.5}\approx7.14 \). So \( t = 7 \) (since \( t\geq5 \) and integer). Cost: \( 2(10)+3.5(7)=20 + 24.5 = 44.5 \leq 45 \), and \( 10 + 7 = 17 \geq 15 \). So (10 drinks, 7 tacos) is a solution.

Step 2: Second Solution

Assume \( d = 15 \), then from \( d + t \geq 15 \), \( t \geq 0 \). Check cost: \( 2(15)+3.5t = 30 + 3.5t \leq 45 \Rightarrow 3.5t \leq 15 \Rightarrow t \leq \frac{15}{3.5}\approx4.29 \). So \( t = 4 \) (integer, \( t\geq0 \)). Cost: \( 2(15)+3.5(4)=30 + 14 = 44 \leq 45 \), and \( 15 + 4 = 19 \geq 15 \). So (15 drinks, 4 tacos) is a solution.

Answer:

Two possible solutions: (10 drinks, 7 tacos) and (15 drinks, 4 tacos). For (10,7): 10+7=17≥15, cost=44.5≤45. For (15,4):15+4=19≥15, cost=44≤45.