Question

How do you solve `1/x+1/2=(5x)/6+1/3`?

Answer 1

`x = 6/5" or " x = -1`

Explanation:

We could treat this as a question on algebraic fractions and add them, but as we are going to solve the equation we don't really want the fractions at all.

Multiply by the LCM of the denominators to cancel them completely. `LCM = color(red)(6x)`

`(color(red)(6cancelx) xx1)/cancelx +(color(red)(cancel6^3x) xx 1)/cancel2 = (color(red)(cancel6x) xx5x)/cancel6 +(color(red)(cancel6^2x) xx1)/cancel3`

`6+3x= 5x^2 +2x" "larr` a quadratic, make `=0`

`5x^2+2x-3x-6 = 0`

`5x^2 -x-6=0`

`(5x-6)(x+1)=0`

Set each factor equal to 0.

`5x -6 =0 " "rarr x = 6/5`

`x+1 = 0" "rarr x =-1`