### Instruction: Find all solutions to the equation \[\frac{\left(\frac{x}{x+1}\right)^2 + 11}{\left(\frac{x}{x+1}\right)^2 + 1} = 2.\]Enter all the solutions, separated by commas.### Response: This equation looks complicated, but I notice that the fraction \(\frac{x}{x+1}\) appears twice, so I will try to simplify it by making a substitution.
Let \(u = \frac{x}{x+1}\).
Then the equation becomes \[\frac{u^2 + 11}{u^2 + 1} = 2.\]
Next, I will try to clear the fractions by multiplying both sides by \(u^2 + 1\).
This gives me \[u^2 + 11 = 2(u^2 + 1).\]
Expanding the right side and moving everything to the left, I get \[u^2 - 2u^2 + 11 - 2 = 0.\]
Simplifying, I get \[-u^2 + 9 = 0.\]
Now I have a quadratic equation in \(u\), which I can try to solve by factoring, completing the square, or using the quadratic formula.
I notice that the equation can be easily factored as \[-(u + 3)(u - 3) = 0.\]
So, the solutions for \(u\) are \(u = -3\) or \(u = 3\).
But I need to find the solutions for \(x\), so I need to substitute back and solve for \(x\).
If \(u = -3\), then \(\frac{x}{x+1} = -3\).
Cross-multiplying, I get \[x = -3x - 3.\]
Adding \(3x\) to both sides, I get \[4x = -3.\]
Dividing by \(4\), I get \[x = -\frac{3}{4}.\]
So, one solution for \(x\) is \(-\frac{3}{4}\).
If \(u = 3\), then \(\frac{x}{x+1} = 3\).
Cross-multiplying, I get \[x = 3x + 3.\]
Subtracting \(3x\) from both sides, I get \[-2x = 3.\]
Dividing by \(-2\), I get \[x = -\frac{3}{2}.\]
So, another solution for \(x\) is \(-\frac{3}{2}\).
Therefore, the two solutions for \(x\) are \(-\frac{3}{4}\) and \(-\frac{3}{2}\).