Five years ago, Tim was twice as old as Jim... Math Question Five years ago, Tim was twice as old as Jim. In three years, Jim’s age will be $2/3$ that of Tim. What are the ages of Tim and Jim now?
 A: hint
Let $ T $ and $ J $ be the ages you want.
Five years ago, we had
$$T-5=2(J-5)$$
In three years, we will have,
$$(J+3)=\frac 23(T+3)$$
Now solve the system.
If you don't find that
$$T=21 \;\; and \;\; J=13$$
It means you made a mistake some minutes ago.
A: Let $t$ be Tim's current age and $j$ be Jim's current age. This is a problem with two equations and two unknowns. The most straightforward way to solve it is to first write out the two equations explicitly, and then solve for one of the variables so you get an equation of the form:
$$ x = \text{some expression not involving $x$} $$
Doing this will allow you to solve for the remaining variable.

Five years ago, Tim was twice as old as Jim.

This is a statement about the ratios of Tim's and Jim's ages five years ago, which we can express directly.
$$ (t-5) = 2 * (j - 5) $$
$$ t - 5 = 2j - 10 $$
$$ t = 2j - 5 $$

In three years Jim's age will be 2/3 that of Tim's.

$$ \frac{j+3}{t+3} = \frac{2}{3} $$
$$ 3j + 9 = 2t + 6 $$
$$ 3j - 2t = -3 $$
We can plug in our earlier equation $ t = 2j - 5 $.
$$ 3j - 2(2j - 5) = -3 $$
$$ 3j - 4j + 10 = -3 $$
$$ -j + 10 = -3 $$
$$ -j = -13 $$
$$ j = 13 $$
Which means $t = 2 * 13 - 5$ or $t = 21$
Let's check the cases
5 years ago
13-5    8
21-5   16

3 years from now
13+3   16
21+3   24

This is indeed our solution.
A: I found the following property of ratios useful for quick calculation: if $\frac{a}{b} = \frac{c}{d}$ then $\frac{a}{b}=\frac{c}{d}=\frac{a+kc}{b+kd}$ as long as $b+kd \ne 0$.
Now,
$T-5=2(J-5), \frac 23 (T+3) = J+3$, we have
$$1=\frac{3J+9}{2T+6} = \frac{2J-10}{T-5} = \frac{3J+9-2(2J-10)}{2T+6-2(T-5)}=\frac{29-J}{16}$$
Then you have $J=13$ immediately, and $T=2(J-5)+5=21.$
