Question

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

Answer 1

`3/(4x+4)`

Explanation:

`(x/(x+1))/(x+x/3)`

First, add the terms in the denominator.

`=(x/(x+1))/((3x)/3+x/3)` (create a common denominator)

`=(x/(x+1))/((4x)/3)` (combine into a single fraction)

`=cancel(x)/(x+1)*3/(4cancel(x))` (rewrite expression and cancel)

`=3/(4(x+1))` (multiply fractions)

`=3/(4x+4)` (distribute in the denominator)

Answer 2

`frac(3 x)(4 (x^(2) + x))`

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

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

`= frac(x)(frac(4)(3) x^(2) + frac(4)(3) x)`

`= frac(x)(frac(4)(3) (x^(2) + x))`

`= frac(3 x)(4 (x^(2) + x))" "larr` factorise

`=frac(3 x)(4x (x + 1))`

`= 3/(4(x+1)`

Answer 3

`3/(4x+4)`

Explanation:

Given:`" "(x/(x+1))/(x+x/3)`

Multiply by 1 but in the form of `3/3` giving:

`(x/(x+1))/(x+x/3)xx3/3" "=" "((3x)/(x+1))/(3x+x)" "=" "((3x)/(x+1))/(4x)`

This is the same as:

`(3x)/(x+1)xx1/(4x)" "=" "x/x xx3/4xx1/(x+1)" "=" "3/(4x+4)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check")`

Set `x=1`

`(x/(x+1))/(x+x/3)" "->" "1/2xx1/((4/3))" "=" "3/8`

`3/(4x+4)" "->" "3/8`

They match!