Question

How do you solve `(4-8x)/(x+1)=8/(x+1)` and find any extraneous solutions?

Answer 1

Solve using the rule `a/b = m/n -> a xx n = bxxm`

Explanation:

`8(x + 1) = (4 - 8x)(x + 1)`

`8x + 8 = 4x - 8x^2 - 8x + 4`

`8x^2 + 12x + 4 = 0`

`4(2x^2 + 3x + 1) = 0`

`4(2x^2 + 2x + x + 1) = 0`

`4(2x(x + 1) + 1(x + 1)) = 0`

`4(2x + 1)(x + 1) = 0`

`x = -1/2 and -1`

Since `x = -1` in the original équation makes the denominator equal to 0, and division by 0 in mathematics is undefined, this is an extraneous solution.

Therefore, our solution set is `{x = -1/2}`.

Hopefully this helps!