Chapter 9.7, Problem 30P

To determine

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

Answer to Problem 30P

The 2-digit number is 37.

Explanation of Solution

Given information:

The two digits in the numerator of a fraction are reversed in its denominator. If 1 is subtracted from both the numerator and the denominator, the value of the resulting fraction is $\frac{1}{2}$ . The fraction whose numerator is the difference and whose denominator is the sum of the units and tens digit equals $\frac{2}{5}$ .

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

  $=\frac{10t+u-1}{10u+t-1}=\frac{1}{2}$

  $\begin{array}{l}=2(10t+u-1)=(10u+t-1)\\ =20t+2u-2=10u+t-1\\ =20t+2u-2-10u-t+1=0\\ =19t-8u-1=0\\ =19t-8u=1\text{ (i)}\end{array}$

The fraction whose numerator is the difference and whose denominator is the sum of the units and tens digit equals $\frac{2}{5}$ .

  $\begin{array}{l}=\frac{u-t}{u+t}=\frac{2}{5}\\ =5(u-t)=2(u+t)\\ =5u-5t=2u+2t\\ \text{= }5u-2u=2t+5t\\ =3u=7t\\ =u=\frac{7}{3}t\text{ (ii) }\end{array}$

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

  $\begin{array}{l}=19t-8\left(\frac{7}{3}\right)t=1\\ =19t-\frac{56}{3}t=1\\ =\frac{57t-56t}{3}=1\\ =\frac{t}{3}=1\\ =t=3\text{ (iii)}\end{array}$

Put equation (iii) in (ii)

  $\begin{array}{l}=u=\frac{7}{3}(3)\\ =u=7\end{array}$

	Tens	Units	Value
Original number	$3$	$7$	$\begin{array}{l}=3(10)+7(1)\\ =30+7\\ =37\end{array}$

Therefore, the 2-digit number is 37.