Chapter 9.7, Problem 29P

To determine

To solve: the system of two equations in two variables.

Answer to Problem 29P

The fraction is $\frac{1}{4}$ .

Explanation of Solution

Given information:

The numerator equals the sum of the two digits in the denominator . The value of the fraction is $\frac{1}{4}$ . When both numerator and denominator are increased by 3, the resulting fraction has the value $\frac{1}{3}$ .

Formula used:

Fraction problem

Step1. The problem asks for .

Step2. Let n = the numerator of the fraction

Let d = the denominator of the fraction.

Then $\frac{n}{d}=\text{the fraction}$ .

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

Step4. Solve the equations.

Calculation:

Let n = the numerator of the fraction.

Let d = the denominator of the fraction.

Therefore, the fraction is

  $\begin{array}{l}=\frac{n}{d}=\frac{1}{4}\\ =4(n)=1(d)\\ =4n=d\text{ (i)}\end{array}$ .

According to the question,

Fraction	increased by 3
Numerator	$n+3$
Denominator	$d+3$

  $\begin{array}{l}=\frac{n+3}{d+3}=\frac{1}{3}\\ =3(n+3)=1(d+3)\\ =3n+9=d\text{+3 }\\ \text{=}3n-d=3-9\\ =3n-d=-6\text{ (ii)}\end{array}$

Putting equation (i) in (ii) we get,

  $\begin{array}{l}=3n-d=-6\\ =3n-(4n)=-6\\ =3n-4n=-6\\ =-n=-6\\ \text{= }n=6\text{ (iii)}\end{array}$

Putting value of n in equation (i)

  $\begin{array}{l}=4(6)=d\\ =24=d\end{array}$

Therefore, the numerator is 6 and the denominator is 24.

The fraction is

  $\begin{array}{l}=\frac{6}{24}\\ =\frac{1}{4}\end{array}$ .

Hence, the original fraction is $\frac{1}{4}.$