Solve for $x$: $\frac{x+1}{x-1} = \frac{x-2}{x+2}$
To solve for $x$, I want to get rid of the fractions by cross-multiplying. That means I multiply the numerator of one fraction by the denominator of the other fraction, and set them equal. So I get $(x+1)(x+2) = (x-1)(x-2)$. Now I want to simplify this equation by expanding the brackets and collecting like terms. I get $x^2 + 3x + 2 = x^2 - 3x + 2$. Subtracting $x^2$ from both sides, I get $3x + 2 = -3x + 2$. Adding $3x$ to both sides, I get $6x + 2 = 2$. Subtracting $2$ from both sides, I get $6x = 0$. Dividing both sides by $6$, I get $x = 0$. To check my answer, I can plug it back into the original equation and see if it makes sense. If $x = 0$, then the left side becomes $\frac{0+1}{0-1} = -1$ and the right side becomes $\frac{0-2}{0+2} = -1$. They are equal, so $x = 0$ is a valid solution. 