Question

How do you solve `6/(x+2)+5/(x-2)=20/(x^2-4)` and check for extraneous solutions?

Answer 1

This equation has no solution.

The "solution" `x=2` that can be obtained by manipulating this equation is extraneous.

Explanation:

Multiply both sides of the equation

`6/{x+2}+5/{x-2} = 20/{x^2-4}`

by `x^2 -4` to get

`6(x-2)+5(x+2) = 20 implies 11x -2 = 20 implies 11x = 22`

This seems to imply that the solution is `x=2`.

Note, however, that the original equation is meaningless for `x=2` (as well as `x=-2`). So, this is an extraneous solution. This does not satisfy the original equation, and, indeed, has arisen because we multiplied both sides of the equation by `x^2-4` - which vanishes when `x=2`.