Question

The difference between two natural numbers is 5 and the difference of their reciprocals is $\dfrac{1}{{10}}$. Find the numbers.

Answer 1

Hint: In this question let the two natural numbers be x and y. Use the constraints of the questions to formulate two different linear equations in two variables. Solve them to get the answer.

Complete step-by-step answer:
Let the first number be x.
And the second number be y.
Let x be greater than y.
Now it is given that the difference of two numbers is 5.
$ \Rightarrow x - y = 5$ ......................... (1)

Now it is also given that the difference of their reciprocals is (1/10).
So as we see difference is positive and (x > y) so ((1/x) < (1/y))
$ \Rightarrow \dfrac{1}{y} - \dfrac{1}{x} = \dfrac{1}{{10}}$

Now simplify this equation we have,
$ \Rightarrow \dfrac{{x - y}}{{xy}} = \dfrac{1}{{10}}$
Now from equation (1) we have,
$ \Rightarrow \dfrac{5}{{xy}} = \dfrac{1}{{10}}$
$ \Rightarrow xy = 50$ .............................. (2)

Now it is a common fact that ${\left( {a + b} \right)^2} = {\left( {a - b} \right)^2} + 4ab$ so use this property we have,
$ \Rightarrow {\left( {x + y} \right)^2} = {\left( {x - y} \right)^2} + 4xy$

Now from equation (1) and (2) we have,
$ \Rightarrow {\left( {x + y} \right)^2} = {\left( 5 \right)^2} + 4\left( {50} \right) = 25 + 200 = 225 = {\left( {15} \right)^2}$
$ \Rightarrow \left( {x + y} \right) = 15$....................... (3)
Now add equation (1) and (3) we have,
$ \Rightarrow x - y + x + y = 5 + 15$
$ \Rightarrow 2x = 20$
Divide by 2 we have,
$ \Rightarrow x = \dfrac{{20}}{2} = 10$

Now substitute this value in equation (1) we have
$ \Rightarrow 10 - y = 5$
$ \Rightarrow y = 10 - 5 = 5$
So the required value of two natural numbers are 10 and 5 respectively.
So this is the required answer.

Note: Natural number is an integer greater than 0. Natural numbers begin at 1 and increment to infinity. The tricky part here was the use of algebraic identity of ${\left( {x + y} \right)^2}$, because the two equations formed directly on solving weren’t giving the values of the variables involved.