Question

Divide $14$ into two parts such that the sum of their reciprocals is $\frac{7}{20}$.

Answer 1

Step 1: Assumption and apply the given condition.

Let the two parts be $x$ and $14-x$.

The reciprocals are as follows:

$\frac{1}{x}\text{and}\frac{1}{14-x}$

Sum of reciprocals of two parts is as follows:

$\frac{1}{x}+\frac{1}{14-x}=\frac{7}{20}$

Simplify the obtained equation as:

$\begin{array}{rcl}\frac{1}{x}+\frac{1}{14-x}& =& \frac{7}{20}\\ \frac{14-x+x}{x\left(14-x\right)}& =& \frac{7}{20}\\ \frac{14}{x\left(14-x\right)}& =& \frac{7}{20}\\ 280& =& 7x\left(14-x\right)\\ 280& =& 98x-7x^2\\ 7x^2-98x+280& =& 0\\ x^2-14x+40& =& 0\end{array}$

Step 2: Factorise the equation.

$\begin{array}{rcl}{x}^2-14x+40& =& 0\\ x^2-10x-4x+40& =& 0\\ x\left(x-10\right)-4\left(x-10\right)& =& 0\\ \left(x-10\right)\left(x-4\right)& =& 0\\ x& =& 10\text{or}4\end{array}$

Step 3: Obtain another part as:

For $x=10$,

$\begin{array}{rcl}14-x& =& 14-10\\ & =& 4\end{array}$

or,

For $x=4$,

$\begin{array}{rcl}14-x& =& 14-4\\ & =& 10\end{array}$

Therefore, the required parts are $4$ and $10$.