Chapter 9.7, Problem 18P

To determine

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

Answer to Problem 18P

The fraction is $\frac{11}{20}$ .

Explanation of Solution

Given information:

If the numerator fraction is increased by 4, the value of the resulting fraction is $\frac{3}{4}$ . If the denominator of the original fraction is increased by 2, the value of the resulting fraction is $\frac{1}{2}$ .

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:

Consider ,

Let d = the denominator of the fraction.

The numerator of a fraction

  $=n=n+4\text{ (i)}$

Therefore, the fraction is $\frac{n+4}{d}$ .

According to the question,

  $\begin{array}{l}=\frac{n+4}{d}=\frac{3}{4}\\ =4(n+4)=3(d)\\ =4n+16=3d\\ \text{=}\frac{1}{3}(4n+16)=d\text{ (ii)}\end{array}$

Fraction	denominator is increased by 2
Numerator	$n$
Denominator	$=d+2$

Therefore,

  $\begin{array}{l}=\frac{n}{d+2}=\frac{1}{2}\\ =2(n)=1(d+2)\\ =2n=d+2\\ =2n-d=2\text{ (iii)}\end{array}$

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

  $\begin{array}{l}=2n-d=2\\ =2n-\frac{1}{3}(4n+16)=2\\ =2n-\frac{4}{3}n-\frac{16}{3}=2\\ =\frac{6n-4n}{3}=2+\frac{16}{3}\\ =\frac{2}{3}n=\frac{6+16}{3}\\ =\frac{2}{3}n=\frac{22}{3}\\ \text{= }n=11\text{ (iv)}\end{array}$

Equation (iv) in (ii)

  $\begin{array}{l}=\frac{1}{3}(4\times 11+16)=d\\ =\frac{1}{3}(44+16)=d\\ =\frac{1}{3}(60)=d\\ =20=d\end{array}$

Therefore, the numerator is 11 and the denominator is 20.

Hence, the fraction is $\frac{11}{20}$ .