Question

The sum of two numbers $a$ and $b$ is $15$, and the sum of their reciprocals $\dfrac{1}{a}$ and $\dfrac{1}{b}$ is $\dfrac{3}{{10}}$.Find the numbers $a$ and $b$.

Answer 1

Hint: Approach the solution by simplify the given conditions i.e. sum of two numbers is $15$ and sum their reciprocals is$\dfrac{3}{{10}}$.

Complete step-by-step answer:
Here the given two numbers are $a$ and $b$.
Given sum of two numbers $a$ and $b$ is$15$ i.e.
$
   \Rightarrow a + b = 15 \\
   \Rightarrow a = 15 - b \to (1) \\
 $
And also given that sum their reciprocals is $\dfrac{3}{{10}}$
$ \Rightarrow \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{3}{{10}}$
By using equation $1$ we can rewrite the ‘$a$’ value as
$ \Rightarrow \dfrac{1}{{b - 15}} - \dfrac{1}{b} = \dfrac{3}{{10}}$
Let us simplify the L.H.S part of the equation by taking the L.C.M of denominators.
$ \Rightarrow \dfrac{{b + 15 - b}}{{b(15 - b)}} = \dfrac{3}{{10}}$ (Here the positive and negative terms of $b$ will get cancelled)
On cross multiplication of the above terms we get
$ \Rightarrow 150 = 3b(15 - b)$
$ \Rightarrow 50 = b(15 - b)$ (Here $3$ cancel $150$ by $50$ times)
Now let us do the further simplification
$
   \Rightarrow 50 = 15b - {b^2} \\
   \Rightarrow {b^2} - 15b + 50 = 0 \\
   \Rightarrow {b^2} - 10b - 5b + 50 = 0 \\
   \Rightarrow b(b - 10) - 5(b - 10) = 0 \\
   \Rightarrow (b - 5)(b - 10) = 0 \\
 $
$\therefore b = 5,b = 10$
Putting both values of $'b'$in equation$(1)$, we get
For $b = 5$
$
   \Rightarrow a = 15 - b \\
   \Rightarrow a = 15 - 5 \\
  \therefore a = 10 \\
 $
For $b = 10$
$
   \Rightarrow a = 15 - b \\
   \Rightarrow a = 15 - 10 \\
  \therefore a = 5 \\
$
So here we got two values $a$ and $b$
Therefore values are a=5, 10 and b=5, 10.

Note: In the above problem we have used equation $(1)$ in condition $2$ to get the b value again from which we can get a value. Simplification is more important.