Question

How do you solve `[ x/(x+1)] = [1/3] + [(x+2)/(4x)]`?

Answer 1

`x=3` or `x= -2/5`

Explanation:

first of all equalize all the denominators as if you're doing a standard fraction addition/subtraction

the right-hand side
`x/(x+1)= (4x*1/(3*4x))+(3*(x+2))/(3*4x))`

`x/(x+1)=(4x+3x+6)/(12x)`

`x/(x+1)=(7x+6)/(12x)`

`x/(x+1)-(7x+6)/(12x)=0`

Now the left-hand side

`((x*12x)/(12x*(x+1)))-((x+1)*(7x+6))/((x+1)*12x)=0`

`(12x^2-(7x^2+6x+7x+6))/(12x*(x+1))=0`

`(12x^2-(7x^2+13x+6))/(12x*(x+1))=0`

`(12x^2-7x^2-13x-6)/(12x*(x+1))=0`

`(5x^2-13x-6)/(12x*(x+1))=0`

the nominator of this fractional equation should be 0 for the expression to yield 0.
So solve for x in:

`(5x^2-13x-6)=0`

Use the quadratic formula (or factorize the expression)

`x=-2/5` or `x= 3`

Feel free to ask questions if you have any