Chapter 9.7, Problem 34P

To determine

To solve: the system of equations in three variables.

Answer to Problem 34P

The age of father is 39 years old now and age of son is 16 years old now.

Explanation of Solution

Given information:

A man is three times as old as his son was at the time when the father was twice as old as his son will be two years from now. The sum of their ages is 55 years.

Formula used:

Age problem.

Step1. The problem asks for the Bob’s age and Claire’s age now.

Step2. Let b = the father’s age now.

Let c = the son‘s age now.

Step3. Use the facts of the problem to write two equations.

Step4. Solve the equations.

Calculation:

Let b = the father’s age now.

Let c = the son‘s age now.

Age	2 years now
Father	$b+2$
Son	$c+2$

According to the equation,

  $=b+c=55\text{ (i)}$

The father age when son’s age is $(c+2)$ ,

  $\begin{array}{l}=b=2(c+2)\\ =b=2c+4\text{ (ii)}\end{array}$

The difference between father and son’s age will be a constant.

  $\begin{array}{l}=b-c=d\\ =c=b-d\text{ (iii)}\end{array}$

Substitute equation (ii) in equation (iii)

  $=c=2c+4-d\text{ (iv)}$

Equation (iv) gives the age of son.

The father’s present age is three times the son’s at that time

  $\begin{array}{l}=b=3(2c+4-d)\\ =b=6c+12-3d\text{ (v)}\\ =b=6c+12-3(b-c)\\ =b=6c+12-3b+3c\\ =b+3b=9c+12\\ =4b=9c+12\text{ (vi)}\end{array}$

Also,

  $\begin{array}{l}=b+c=55\text{ (vii)}\\ \text{Multiply by 2}\\ =2b+2c=110\\ \text{Subtract the equation by (vi)}\\ =\frac{13c}{13}=\frac{208}{13}\\ =c=16\end{array}$

Putting value of c in equation (vii) we get,

  $\begin{array}{l}=b+16=55\\ =b=55-16\\ =b=39\end{array}$

Hence, the age of father is 39 years old now and age of son is 16 years old now.