Question

How do you solve `1 + 8/(x - 5) = 3/x` and find any extraneous solutions?

Answer 1

There are no Real solutions, but there are Complex solutions:

`x = +-sqrt(15)i`

Explanation:

Given:

`1+8/(x-5) = 3/x`

Note that neither `0` nor `5` can be a solution since they result in division by `0`. So if our derived equations have solutions `0` or `5` then they are extraneous.

Multiply both sides by `x` (possibly introducing an extraneous solution `0`) to get:

`x + (8x)/(x-5) = 3`

Multiply both sides by `(x-5)` (possibly introducing an extraneous solution `5`) to get:

`x(x-5) + 8x = 3(x-5)`

Expand both sides to get:

`x^2-5x+8x = 3x-15`

which simplifies to:

`x^2+3x = 3x-15`

Subtract `3x` from both sides to get:

`x^2 = -15`

This has no Real solutions since `x^2 >= 0` for any Real value of `x`.

If we are interested in Complex solutions, then add `15` to both sides and transpose to get:

`0 = x^2+15 = x^2 - (sqrt(15)i)^2 = (x-sqrt(15)i)(x+sqrt(15)i)`

where `i` is the imaginary unit, with the property that `i^2 = -1`.

Hence solutions `x = +-sqrt(15)i`