## Sunday, May 20, 2018

### Algebra 2 Problems of the Day (open ended)

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

August 2017, Part III

All Questions in Part III are worth 4 credits. Work need be shown (or explained or justified) for full credit. Correct numerical answers with no work receive one credit.

33. Solve for all values of p:

3p / (p - 5) - 2 / (p + 3) = p / (p + 3)

Move the fraction with the (p + 3) denominator to the right side and combine the fractions. Next cross-multiply to get a quadratic equation. Set the equation equal to 0 by moving everything to the left side. Divide the equation by 2 to get rid of the leading coefficient. The remaining trinomial is ridiculously easy to factor into (p + 5) and (p - 1). That means that the solutions are p = -5 and p = -1.
According to original equation, +5 and -3 would have to be rejected as solutions. However, -5 is fine, so do not reject it (or -1, for that matter).

Refer to the image below:

34. Simon lost his library card and has an overdue library book. When the book was 5 days late, he owed \$2.25 to replace his library card and pay the fine for the overdue book. When the book was 21 days late, he owed \$6.25 to replace his library card and pay the fine for the overdue book.
Suppose the total amount Simon owes when the book is n days late can be determined by an arithmetic sequence. Determine a formula for an, the nth term of this sequence.

The difference between the 5th term and the 21st term is \$4.00 over 21-5 = 16 days. That's a constant different of \$.25 per day.
Using the formula an = a1 + (n-1)d, we can find the fine for the first day
2.25 = a1 + 4(.25)
2.25 = a1 + 1.00
1.25 = a1 is the first day.
an = 1.25 + (n - 1)(.25)

To find the cost on the 60th day, substitute 60 for n:
an = 1.25 + (60 - 1)(.25) = 16.00
The fine will be \$16.00