Question

How do you solve the rational equation `(x+1)/(x-1)=x/3 + 2/(x-1)`?

Answer 1

`x = 3`

Explanation:

Here we go...

`(x+1)/(x-1)` = `x/3` + `2/(x-1)`

`(x+1)/(x-1)` = `(x^2 - x + 6)/(3x - 3)`

`(x + 1)(3x - 3)` = `(x-1)(x^2 - x + 6)`

`3x^2 - 3` = `x^3 - 2x^2 + 7x - 6`

`x^3 - 5x^2 + 7x - 3 = 0`

By factorizing this equation we get :

`(x-1)(x-1)(x-3)` = 0

`:.` `x = 1 or 3`

But as observed in the given question `x - 1` is in denominator, So `x` cannot be equal to 1.

Hence, `x = 3`

Answer 2

`x=3`

Explanation:

We have an equation with fractions.

It is possible to get rid of the denominators immediately by multiplying each term by the LCM of the denominators, which in this case is `color(blue)(3(x-1))`

`(color(blue)(3cancel((x-1)))xx(x+1))/cancel((x-1))=(color(blue)(cancel3(x-1))xx x)/cancel3 + (2color(blue)(xx3cancel((x-1))))/cancel((x-1))`

This leaves us with:

`3(x+1) = (x-1)x +6" "larr` no fractions

`3x+3 = x^2-x+6" "larr` remove the brackets

`0 = x^2-4x+3" "larr` make the quadratic equal to `0`

`(x-3)(x-1)=0" "larr` factorise

If `x-3 = 0," " rArr x=3`

If `x-1=0," "rArr x = 1`

However, `x=1` is an extraneous solution.
In this equation, `x !=1` because that will make the denominators `0`.

Therefore there is only one solution.

`x =3`