I was asked to help my sister with a bit of precalculus homework and completely drew a blank upon encountering this problem.
I believe it was asking to "balance the equation, and set the answer to zero."
This IS the equation
(x+1)^-1=(x^-1)+1
The answers were given. I BELIEVE they were 4 and 1.3333 (though I'm not 100% certain on that.)
I'm interested in the journey however, not the destination.
I would greatly appreciate any help.
So we want to solve
$$\frac{1}{x + 1} = \frac{1}{x} + 1$$
Multiplying throughout by $x(x+1) = x^2 + x$,
$$x = (x + 1) + x^2 + x$$
$$x = x^2 + 2x + 1$$ $$x^2 + x + 1 = 0$$
Unfortunately, this equation does not have any real solutions for $x$. This is because
$$\begin{align}x^2 + x + 1 &= \left(x + \frac{1}{2}\right)^2 + 1 - \frac{1}{4}\\ &=\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\\ &\ge \frac{3}{4} > 0\end{align}$$
for all real values of $x$. That is to say, the $x^2 + x + 1$ is always greater than $0$. You can consult the Wolf directly to verify that no real solutions exist.