The problem asks for an inequality to represent how many $20 bills you can withdraw from a checking account without going below the minimum balance of $50. You currently have $280 in the account.

Let's define the number of $20 bills you can withdraw as $x$. Each $20 bill reduces the account balance by $20, so the amount you withdraw is $20x$.

Inequality Setup:

You want to make sure that after withdrawing money, your balance is at least $50. The inequality for this is:

$280 - 20x \geq 50$

Now, let's solve this inequality:

Solving the Inequality:

1. Subtract 50 from both sides: $280 - 20x \geq 50$

   Subtract 50 from both sides: $280 - 20x \geq 50$ $230 \geq 20x$

2. Divide both sides by 20: $\frac{230}{20} \geq x$ $11.5 \geq x$

Since $x$ must be an integer (you can't withdraw half a $20 bill), the maximum number of $20 bills you can withdraw is 11.

Final Answer:

You can withdraw up to 11 $20 bills without going below the minimum balance of $50.

Would you like further details, or do you have any questions?

Here are 5 related questions for practice:

1. What would happen if the balance was $300 instead of $280?
2. How would the inequality change if you could withdraw $50 bills instead?
3. What if the minimum balance requirement was $100 instead of $50?
4. How would you adjust the inequality if you had $10 bills instead of $20 bills?
5. How does increasing the minimum balance affect the number of $20 bills you can withdraw?

Tip: Always check whether your answer makes sense in the context of the problem. In this case, since you can't withdraw a fraction of a $20 bill, it's important to round down to the nearest whole number.