# How do you simplify ((x)/(x+1))/(x+x/3)?

Jul 13, 2017

$\frac{3}{4 x + 4}$

#### Explanation:

$\frac{\frac{x}{x + 1}}{x + \frac{x}{3}}$

First, add the terms in the denominator.

$= \frac{\frac{x}{x + 1}}{\frac{3 x}{3} + \frac{x}{3}}$ (create a common denominator)

$= \frac{\frac{x}{x + 1}}{\frac{4 x}{3}}$ (combine into a single fraction)

$= \frac{\cancel{x}}{x + 1} \cdot \frac{3}{4 \cancel{x}}$ (rewrite expression and cancel)

$= \frac{3}{4 \left(x + 1\right)}$ (multiply fractions)

$= \frac{3}{4 x + 4}$ (distribute in the denominator)

Jul 13, 2017

$\frac{3 x}{4 \left({x}^{2} + x\right)}$

= 3/(4(x+1)

#### Explanation:

We have: $\frac{\frac{x}{x + 1}}{x + \frac{x}{3}}$

$= \frac{x}{x + 1} \times \frac{1}{x + \frac{x}{3}}$

$= \frac{x}{\left(x + 1\right) \left(x + \frac{x}{3}\right)}$

= frac(x)((x)(x) + (x)(frac(x)(3)) + (1)(x) + (1)(frac(x)(3))

$= \frac{x}{{x}^{2} + \frac{{x}^{2}}{3} + x + \frac{x}{3}}$

$= \frac{x}{{x}^{2} \left(1 + \frac{1}{3}\right) + x \left(1 + \frac{1}{3}\right)}$

$= \frac{x}{\frac{4}{3} {x}^{2} + \frac{4}{3} x}$

$= \frac{x}{\frac{4}{3} \left({x}^{2} + x\right)}$

$= \frac{3 x}{4 \left({x}^{2} + x\right)} \text{ } \leftarrow$ factorise

$= \frac{3 x}{4 x \left(x + 1\right)}$

= 3/(4(x+1)

Jul 13, 2017

$\frac{3}{4 x + 4}$

#### Explanation:

Given:$\text{ } \frac{\frac{x}{x + 1}}{x + \frac{x}{3}}$

Multiply by 1 but in the form of $\frac{3}{3}$ giving:

$\frac{\frac{x}{x + 1}}{x + \frac{x}{3}} \times \frac{3}{3} \text{ "=" "((3x)/(x+1))/(3x+x)" "=" } \frac{\frac{3 x}{x + 1}}{4 x}$

This is the same as:

$\frac{3 x}{x + 1} \times \frac{1}{4 x} \text{ "=" "x/x xx3/4xx1/(x+1)" "=" } \frac{3}{4 x + 4}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Check}}$

Set $x = 1$

$\frac{\frac{x}{x + 1}}{x + \frac{x}{3}} \text{ "->" "1/2xx1/((4/3))" "=" } \frac{3}{8}$

$\frac{3}{4 x + 4} \text{ "->" } \frac{3}{8}$

They match!