Anne, Barbara, and Cindy are three sisters who were born in different years. Cindy is 5 years younger than Barbara, and Anne is 8 years older than Cindy. Anne is now a years old.
It is a wonderful age. . . .
Using a as a variable, write algebraic expressions describing each part of problems 1 and 2. Simplify if possible.

1. 1. Cindy’s age now
   2. Barbara’s age now
   3. The sum of Barbara’s and Cindy’s ages now
   4. The sum of Barbara’s and Cindy’s ages in three years
   5. The sum of Barbara’s and Cindy’s ages k years from now
   6. The sum of Barbara’s and Cindy’s ages k years ago
      Assume that both sisters had been born at the time. Read each problem carefully: the words can be tricky!
      
2. 1. The difference between Barbara’s and Cindy’s ages now
   2. The difference between Barbara’s and Cindy’s ages two years ago
      Both had been born already!
      
   3. The difference between Barbara and Cindy’s ages two years from now
   4. The difference between Barbara’s and Cindy’s ages k years from now
3. Write the verbal description (or several) for the algebraic expressions below. For instance: (a - 3) means “Alice’s age three years ago” or “Cindy’s age in five years.”
   1. 3a
   2. 3a - 11
   3. 2a - 9
   4. a(a - 3) + (a + 3)

There are no hints for this problem sequence.

1. 1. a - 8
   2. a - 3
   3. 2a - 11
   4. 2a - 5
   5. 2a + 9
   6. 2a - 11 + 2k
   7. 2a - 11 - 2k
2. 1. 5
   2. 5
   3. 5
   4. 5
3. See solutions.

1. 1. Since Cindy is 8 years younger than Anne (who is age a), Cindy’s age is a - 8.
   2. Since Barbara is 5 years older than Cindy (who is age a - 8), Barbara’s age is (a - 8) + 5, or a - 3. Other way to think of this is to realize if Barbara is 5 years older than Cindy and Anne is 8 years older than Cindy, then Anne is 3 years older than Barbara.
   3. Adding the results from parts (a) and (b) gives a-8+ a - 3, which is 2a - 11.
   4. In three years, Cindy will be age a - 8 + 3, which is a - 5. Barbara will be age a - 3 + 3, which is a. The sum of their ages will be a - 5 + a, which is 2a - 5.
   5. Ten years from now, Cindy and Barbara will be a-8+ 10 and a-3+10. The sum is (a-8+10)+(a-3+10), which is 2a + 9.
   6. In k years Barbara and Cindy will be a - 3 + k and a-8+k, so the sum of their ages is a-3+k+a-8+k, which is 2a - 11 + 2k.
   7. Barbara and Cindy were a-3-k and a-8-k, k years ago. The sum of their ages then was a-3-k+a-8-k, which is 2a - 11 - 2k.
2. Barbara’s older, so subtract Cindy’s age from Barbara’s in each case. These could all be done by remembering that Barbara is 5 years older than Cindy. The difference between their ages will always be 5. This can also be found by subtracting algebraic expressions, as shown below.
   1. The difference now is (a - 3) - (a - 8). Simplify:
   2. The difference two years ago was (a-3-2)-(a-8-2). Simplify:
   3. The difference two years from now will be (a-3-2)- (a - 8 - 2).
   4. The difference k years from now will be (a - 3 + k) - (a - 8 + k).
3. There are many possible answers; examples are given.
   1. Anne’s age tripled, Anne’s age 2a years from now, Barbara’s age 2a + 3 years from now
   2. The sum of the ages of all three sisters now
   3. The sum of Barbara’s and Cindy’s ages a year from now
   4. The product of Anne’s and Barbara’s ages now plus Anne’s age three years from now