Chapter 9.7, Problem 31P

To determine

To solve: the system of equations in two variables.

Answer to Problem 31P

The age of Laura is 18 years old now and age of Maria is 12 years old now.

Explanation of Solution

Given information:

Laura is 3 times as old as Maria was when Laura was as old as Maria is now. In 2years, Laura will be twice as old as Maria was 2 years ago.

Formula used:

Age problem.

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

Step2. Let b = the Laura’s age .

Let c = the Maria‘s age .

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

Step4. Solve the equations.

Calculation:

Consider ,

Let b = the Laura’s age .

Let c = the Maria’s age .

Let the age of Laura and Maria in Y years ago be :-

Age	Y year ago
Laura	$b-Y$
Maria	$c-Y$

According to the equation,

  $=b=3(c-Y)\text{ (i)}$

Also,

  $\begin{array}{l}=b-Y=c\\ =Y=b-c\text{ (ii)}\end{array}$

Substituting equation (ii) in (i)

  $\begin{array}{l}=b=3\left(c-(b-c)\right)\\ =b=3(c-b+c)\\ =b=3(2c-b)\\ =b=6c-3b\\ =b+3b-6c=0\\ =4b-6c=0\\ =2(2b-3c)=0\\ =2b-3c=0\text{ (iii)}\end{array}$

  $=b=\frac{3}{2}c\text{ (iv)}$

Now,

Age	2 year	2 years ago
Laura	$b+2$	$b-2$
Maria	$c+2$	$c-2$

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

Putting equation (iv) in (v) we get,

  $\begin{array}{l}=\frac{3}{2}c-2c=-6\\ =\frac{3c-4c}{2}=-6\\ =-c=(-6)(2)\\ =c=12\end{array}$

Putting value of c in equation (iv)

  $\begin{array}{l}=b=\frac{3}{2}(12)\\ =b=3(6)\\ =b=18\end{array}$

Hence, the age of Laura is 18 years old now and age of Maria is 12 years old now .