Question

The sum of Joe's and Sheila's ages is 115. Fourteen years ago, Joe was twice as old as Sheila. How old is Sheila now?

Answer 1

Answer:  Sheila today = $46\dfrac{1}{3}$  yrs old

Step-by-step explanation:

   J + S = 115 v⇒  J = 115 - S

Current Ages                        Ages 14 years ago

 Joe (J) = 115 - S                      J - 14 = 2(S - 14)

 Sheila (S) = S

Substitute J = 115 - S into the "14 years ago" equation

      J    - 14 = 2(S - 14)

(115 - S) - 14 = 2(S - 14)

111 -S = 2S - 28

111 = 3S  - 28

139 = 3S

46 $\frac{1}{3}$ = S

It is odd that the result was not an integer.  I wonder if you meant to type "Joe was twice as old as Sheila is today.  That would change the equation to:

J - 14 = 2S

111 - S = 2S

111 = 3S

37 = S