Answer:
a.
[tex]280 - n*20\geq50\\n\leq11[/tex]
b.
[tex]280 - n*(20+2.50)\geq 50\\n\leq10[/tex]
Explanation:
a. For the first situation, the goal is to make sure that a minimum amount of $50 (balance ≥ 50) is left on the account after withdrawing "n" $20 bills from an initial $280 (balance = 280 - 20n). The inequality can be modeled and solved as follows:
[tex]280 - n*20\geq50\\n\leq\frac{280-50}{20} \\n\leq11.5[/tex]
Since the problem deals with $20 bills, only whole quantities should be considered and then, the inequality should be written as:
[tex] n\leq11[/tex]
b. This situation is very similar to the previous one but it has an additional $2.50 fee for each $20 bill withdrawn (balance = 280 - n(20+2.50)). The inequality can be modeled as follows
[tex]280 - n*(20+2.50)\geq 50\\n\leq\frac{280-50}{22.5}\\n\leq10.22[/tex]
Once again, only whole quantities are considered, so the inequality yields:
[tex]n\leq10[/tex]