Chapter 9.7, Problem 33P

To determine

To solve: the system of equations in three variables.

Answer to Problem 33P

The age of Cindy is 26 years old now , age of Paul is 18 years old now and age of Sue is 8 years old now.

Explanation of Solution

Given information:

Cindy’s equals the sum of Paul’s age and Sue’s age. Two years ago, Cindy was 4 times as old as Sue was, and two years from now, Cindy will be 1.4 times as old as Paul will be.

Formula used:

Age problem.

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

Step2. Let b = the Cindy’s age now.

Let c = the Paul ‘s age now.

Let d = the Sue ‘s age now.

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

Step4. Solve the equations.

Calculation:

Let b = the Cindy’s age now.

Let c = the Paul ‘s age now.

Let d = the Sue ‘s age now.

Age	2 year ago	2 years now
Cindy	$b-2$	$b+2$
Paul	$c-2$	$c+2$
Sue	$d-2$	$d+2$

According to the equation,

  $=b=c+d\text{ (i)}$

Also,

  $\begin{array}{l}=b-2=4(d-2)\\ =b-2=4d-8\\ =b=4d-8+2\\ =b=4d-6\text{ (ii)}\end{array}$

And

  $\begin{array}{l}=b+2=1.4(c+2)\\ =b+2=1.4c+2.8\\ =b=1.4c+2.8-2\\ =b=1.4c+0.8\text{ (iii)}\end{array}$

Putting value of b in equation (ii) and (iii) we get,

  $\begin{array}{l}=c+d=4d-6\\ =c+d-4d=-6\\ =c-3d=-6\text{ (iv)}\end{array}$

  $\begin{array}{l}=c+d=1.4c+0.8\\ =c+d-1.4c=0.8\\ =-0.4c+d=0.8\text{ (v)}\end{array}$

Now , solving equation (iv) and (v) we get,

Multiplying equation (v) by 3 and adding equation with equation (iv)

  $\begin{array}{l}=c-1.2c=-6+2.4\\ =-0.2c=-3.6\\ =\frac{0.2c}{0.2}=\frac{3.6}{0.2}\\ =c=18\end{array}$

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

  $\begin{array}{l}=-0.4(18)+d=0.8\\ =-7.2+d=0.8\\ =d=0.8+7.2\\ =d=8\end{array}$

Putting value of c and d in equation (i) we get,

  $\begin{array}{l}=b=8+18\\ =b=26\end{array}$

Age	Present age
Cindy	26
Paul	18
Sue	8

Hence, the age of Cindy is 26 years old now , age of Paul is 18 years old now and age of Sue is 8 years old now.