Question

Zyan’s apartment building has a washing machine that uses a card for payment. There is $50 on the card right now, and her family keeps a minimum balance of at least $10 on the card. Each load of laundry costs $1.50. How many loads of laundry can Zyan’s family do and still have at least $10 on the card?
Solve the associated equation. Show your work or solve on paper. Write your answer as a fraction.
50 – 1.5x ≥ 10
50−1.5x
=
10

x
=

Answer 1

To determine how many loads of laundry Zyan's family can do while keeping at least $10 on their card, we can start with the inequality:

\[ 50 - 1.5x \geq 10 \]

Where:
- \( 50 \) is the initial balance on the card.
- \( 1.5 \) is the cost of each load of laundry.
- \( x \) is the number of loads of laundry.

Now, we can solve the inequality step by step.

1. Subtract \( 50 \) from both sides:

\[
-1.5x \geq 10 - 50
\]

\[
-1.5x \geq -40
\]

2. Divide both sides by \( -1.5 \). Remember that when you divide or multiply an inequality by a negative number, you must flip the inequality sign:

\[
x \leq \frac{-40}{-1.5}
\]

3. Now we calculate \( \frac{-40}{-1.5} \):

\[
x \leq \frac{40}{1.5}
\]

To simplify \( \frac{40}{1.5} \):

\[
\frac{40}{1.5} = \frac{40 \times 10}{1.5 \times 10} = \frac{400}{15}
\]

Now simplify \( \frac{400}{15} \):

\[
\frac{400}{15} = \frac{80}{3}
\]

So, we have:

\[
x \leq \frac{80}{3}
\]

Therefore, the answer is that Zyan’s family can do at most \( \frac{80}{3} \) loads of laundry while keeping at least $10 on the card.

So:
\[
x = \frac{80}{3} \approx 26.67
\]

Since they can’t do a fraction of a load, they can do a maximum of \( 26 \) loads and still keep at least $10 on the card.