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

3 Answers
Jul 13, 2017

Answer:

#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:

#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)#

Jul 13, 2017

Answer:

#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!