# How do you simplify (x+1)/x - x/(x+1)?

May 18, 2018

(2x+1)/(x(x+1)

#### Explanation:

Find the lowest common factor which is $x \left(x + 1\right)$

$\frac{x + 1}{x} = \frac{x}{x + 1}$

$\frac{\left(x + 1\right) \left(x + 1\right) - \left(x \times x\right)}{x \left(x + 1\right)}$

(x^2+x+x+1 - x^2)/(x(x+1)

(cancel(x^2-x^2)+2x+1)/(x(x+1)

(2x+1)/(x(x+1)

May 18, 2018

$\frac{2 x + 1}{x \left(x + 1\right)}$

#### Explanation:

$\text{we require the fractions to have a "color(blue)"common denominator}$

$\text{to obtain this}$

$\text{multiply the numerator/denominator of } \frac{x + 1}{x}$
$\text{by } \left(x + 1\right)$

$\text{and multiply numerator/denominator of "x/(x+1)" by } x$

$= \frac{\left(x + 1\right) \left(x + 1\right)}{x \left(x + 1\right)} - {x}^{2} / \left(x \left(x + 1\right)\right)$

$\text{the fractions now have a common denominator so expand}$
$\text{and subtract the numerator leaving the denominator}$

$= \frac{\cancel{{x}^{2}} + 2 x + 1 \cancel{- {x}^{2}}}{x \left(x + 1\right)}$

$= \frac{2 x + 1}{x \left(x + 1\right)}$

May 18, 2018

$\frac{x + 1}{x} - \frac{x}{x + 1} = \frac{1}{x} + \frac{1}{x + 1}$

#### Explanation:

I am not sure what is meant by simplified here, but we can find:

$\frac{x + 1}{x} - \frac{x}{x + 1} = \frac{x + 1}{x} - \left(\frac{x + 1 - 1}{x + 1}\right)$

$\textcolor{w h i t e}{\frac{x + 1}{x} - \frac{x}{x + 1}} = \left(1 + \frac{1}{x}\right) - \left(1 - \frac{1}{x + 1}\right)$

$\textcolor{w h i t e}{\frac{x + 1}{x} - \frac{x}{x + 1}} = \frac{1}{x} + \frac{1}{x + 1}$