Solving a Medium-Level Algebra Problem: Father's Age Explained

[Jameson]

With some students I've tutored lately we've come across quite a bit of word problems that can be a headache to set up, so for all of you younger high school students here is a common medium-level of difficulty problem you might see in algebra class or on college entrance exams.

A father is three times as old as his son. After 15 years the father will be twice as old as his son’s age at that time. What is the present age of the father?

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[Jameson]

Congratulations to the following members for their correct solutions:

1) Sudharaka
2) hp12345
3) Reckoner
4) Siron
5) soroban
6) BAdhi
7) veronica1999

Solution (from soroban):
[sp]Let x = son's age now.
Then 3x = father's age now.

I construct a table for this type of problem.
Make one row for each person.

\begin{array}{|c|c|}\hline \text{Father} & 3x \\ \hline \text{Son} & x \\ \hline \end{array}Then I make two columns.
. . [/color]The first is "Now" for the present ages;
. . [/color]the other is "Then" for the ages at some other time (past or future).

\begin{array}{|c|c|c|} & \text{Now} & \text{Then} \\ \hline \text{Father} & 3x & \\ \hline \text{Son} & x & \\ \hline \end{array}This time "Then" is 15 years in the future.
The father will be 15 years older: 3x + 15
The son will also be 15 years older: x + 15

\begin{array}{|c|c|c|} & \text{Now} & \text{Then} \\ \hline \text{Father} & 3x & 3x+15 \\ \hline \text{Son} & x & x + 15\\ \hline \end{array}The equation comes from the last column.

. . [/color]\underbrace{\text{Father's age (then)}} \;=\;2\times\underbrace{\text{Son's age (then)}}

. . . . . . . [/color]3x+15 \qquad\;\;=\;2 \quad\times\quad (x + 15)And there is our equation! . . . 3x+15 \;=\;2(x+15)Solve for x\!:\;\begin{Bmatrix}x &=& 15 & \text{Son} \\ 3x &=& 45 & \text{Father} \end{Bmatrix}[/sp]