Chapter 9.7, Problem 28P

To determine

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

Answer to Problem 28P

The fraction is $\frac{12}{21}$ .

Explanation of Solution

Given information:

The two digits in the numerator of a fraction whose value is $\frac{4}{7}$ are reversed in its denominator. The reciprocal of the fraction is the value obtained when 16 is added to the original numerator and 5 is subtracted from the original denominator.

Formula used:

Digit problems are based on our decimal system of numeration.

Step1. The problem asks for the original number.

Step2. Let t = the tens digit of the original number.

Let u = the units digit of the original number.

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

Step4. Solve the equations.

Calculation:

Let t = the tens digit of the original number.

Let u = the units digit of the original number.

	Tens	Units	Value
Original number	t	u	$10t+u$
Number reversed	u	t	$10u+t$

This implies

  $=\frac{10t+u}{10u+t}=\frac{4}{7}\text{ (i)}$

  $\begin{array}{l}=7(10t+u)=4(10u+t)\\ =70t+7u=40u+4t\\ =70t+7u-40u-4t=0\\ =66t-33u=0\\ =2t-u=0\\ \text{=}2t=u\text{ (ii)}\end{array}$

The reciprocal of the fraction is the value obtained when 16 is added to the original numerator and 5 is subtracted from the original denominator.

  $\begin{array}{l}=\frac{10t+u+16}{10u+t-5}=\frac{7}{4}\\ =4(10t+u+16)=7(10u+t-5)\\ =40t+4u+64=70u+7t-35\\ \text{= }40t+4u-70u-7t=-35-64\\ =33t-66u=-99\\ =\frac{33(t-2u)}{33}=-\frac{99}{33}\\ =t-2u=-3\text{ (iii) }\end{array}$

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

  $\begin{array}{l}=t-2(2t)=-3\\ =t-4t=-3\\ =-3t=-3\\ =\frac{3t}{3}=\frac{3}{3}\\ =t=1\text{ (iv)}\end{array}$

Put equation (iv) in (ii)

  $\begin{array}{l}=u=2(1)\\ =u=2\end{array}$

Therefore,

  $\begin{array}{l}=\frac{10(1)+2}{10(2)+1}\\ =\frac{10+2}{20+1}\\ =\frac{12}{21}\end{array}$

Hence the original fraction is $\frac{12}{21}$ .