Question

on a family outing, tyler bought 5 cups of hot cocoa and 4 pretzels for $18.40. some of his family members would like a second serving, so he went back to the same food stand and bought another 2 cups of hot cocoa and 4 pretzels for $11.20. here is a system of equations that represent the quantities and constraints in this situation. $5c + 4p = 18.40$ $2c + 4p = 11.20$ 1. what is the solution to the system of equations? 2. what does the solution to the system, $(c, p)$, represent in this situation?

Explanation:

Step1: Subtract equations to eliminate $p$

$$(5c + 4p) - (2c + 4p) = 18.40 - 11.20$$
$$3c = 7.20$$

Step2: Solve for $c$

$$c = \frac{7.20}{3} = 2.40$$

Step3: Substitute $c$ into second equation

$$2(2.40) + 4p = 11.20$$
$$4.80 + 4p = 11.20$$

Step4: Solve for $p$

$$4p = 11.20 - 4.80 = 6.40$$
$$p = \frac{6.40}{4} = 1.60$$

The ordered pair $(c, p)$ corresponds to the cost of each item in the scenario. $c$ is the variable assigned to hot cocoa, and $p$ is assigned to pretzels, so their values represent the individual prices of these two items.

Answer:

for Question 1:
$c = 2.40$, $p = 1.60$

---