Question

How do you simplify $\dfrac{\dfrac{x}{y}-\dfrac{y}{x}}{\dfrac{1}{y}+\dfrac{1}{x}}$ and can you use a calculator to simplify?

Answer 1

Hint: First try to simplify by taking the numerator and the denominator separately. In both the cases take the l.c.m. as $x.y$. After simplifying again write the expression by taking the simplified form of the numerator and the denominator together. Cancel out $x.y$ both from the numerator and the denominator and apply ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ formula for ${{x}^{2}}-{{y}^{2}}$. Again cancel out $\left( x+y \right)$ both from the numerator and the denominator to get the required result.

Complete step by step answer:
 The expression we have, $\dfrac{\dfrac{x}{y}-\dfrac{y}{x}}{\dfrac{1}{y}+\dfrac{1}{x}}$
It’s numerator can be simplified as
$\dfrac{x}{y}-\dfrac{y}{x}$
As we know the l.c.m. of ‘x’ and ‘y’ is $x.y$
So, taking l.c.m. as $x.y$, we get
$\begin{align}
  & \Rightarrow \dfrac{x\cdot x-y\cdot y}{x\cdot y} \\
 & \Rightarrow \dfrac{{{x}^{2}}-{{y}^{2}}}{x.y} \\
\end{align}$
It’s denominator can be simplified as
$\dfrac{1}{y}+\dfrac{1}{x}$
Again, taking the l.c.m. of ‘x’ and ‘y’ as $x.y$, we get
$\begin{align}
  & \Rightarrow \dfrac{1\cdot x+1.y}{x\cdot y} \\
 & \Rightarrow \dfrac{x+y}{x\cdot y} \\
\end{align}$
Considering the simplified forms of both the numerator and the denominator together, the expression can be written as
$\dfrac{\dfrac{{{x}^{2}}-{{y}^{2}}}{x.y}}{\dfrac{x+y}{x\cdot y}}$
Which can also be written as
$\Rightarrow \dfrac{{{x}^{2}}-{{y}^{2}}}{x.y}\times \dfrac{x.y}{x+y}$
Cancelling $x.y$ from both numerator and denominator, we get
$\Rightarrow \dfrac{{{x}^{2}}-{{y}^{2}}}{x+y}$
Again, we know ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
So, ${{x}^{2}}-{{y}^{2}}$ can be written as ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$
Now, our expression becomes
$\Rightarrow \dfrac{\left( x+y \right)\left( x-y \right)}{x+y}$
Again cancelling $\left( x+y \right)$ both from the numerator and the denominator, we get
$\Rightarrow x-y$
Hence, $\dfrac{\dfrac{x}{y}-\dfrac{y}{x}}{\dfrac{1}{y}+\dfrac{1}{x}}=x-y$
We can’t use calculators for algebraic expression.

Note: The numerator and the denominator of the given expression should be simplified by taking each of them separately to avoid any error or confusion. The expression $\dfrac{\dfrac{{{x}^{2}}-{{y}^{2}}}{x.y}}{\dfrac{x+y}{x\cdot y}}$ means $\dfrac{{{x}^{2}}-{{y}^{2}}}{x.y}\div \dfrac{x+y}{x\cdot y}$, so by taking the reciprocal of $\dfrac{x+y}{x\cdot y}$ it can be written as $\dfrac{{{x}^{2}}-{{y}^{2}}}{x.y}\times \dfrac{x.y}{x+y}$.