Question

How do you simplify $\frac{{\displaystyle \frac{x}{9}}-{\displaystyle \frac{1}{x}}}{1+{\displaystyle \frac{3}{x}}}$?

Answer 1

Step-1: Simplify the exoression:

Algebraic property $\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$

Use this property to simplify:

$\frac{{\displaystyle \frac{x}{9}}-{\displaystyle \frac{1}{x}}}{1+{\displaystyle \frac{3}{x}}}=\frac{{\displaystyle \frac{x^2-9}{9x}}}{{\displaystyle \frac{x+3}{x}}}$

Multiplying numerator and denominator by the reciprocal of $\frac{x+3}{x}$.

$\begin{array}{rcl}\frac{{\displaystyle \frac{x}{9}-\frac{1}{x}}}{1+{\displaystyle \frac{3}{x}}}& =& \frac{\left({\displaystyle \frac{x^2-9}{9x}}\right)\times \left({\displaystyle \frac{x}{x+3}}\right)}{\left({\displaystyle \frac{x+3}{x}}\right)\times \left({\displaystyle \frac{x}{x+3}}\right)}\\ & \Rightarrow & \frac{{\displaystyle \frac{x}{9}-\frac{1}{x}}}{1+{\displaystyle \frac{3}{x}}}=\frac{x(x^2-9)}{9x(x+3)}\\ & \Rightarrow & \frac{{\displaystyle \frac{x}{9}-\frac{1}{x}}}{1+{\displaystyle \frac{3}{x}}}=\frac{(x^2-9)}{9(x+3)}\end{array}$

Step-2: Factorize the expression:

Factor the numerator and cancel the like terms:

$\begin{array}{rcl}\frac{(x^2-9)}{9(x+3)}& =& \frac{(x-3)\times (x+3)}{(x+3)\times 9}[\mathrm{using},a^2-b^2=(a+b)\times (a-b\left)\right]\\ & =& \frac{x-3}{9}\end{array}$

Hence, the simplified expression is $\frac{\mathbf{x}\mathbf{-}\mathbf{3}}{\mathbf{9}}\mathbf{.}$