### Instruction: Let $x_1,$ $x_2,$ $\dots,$ $x_{100}$ be real numbers such that $x_1 + x_2 + \dots + x_{100} = 1$ and
\[\frac{x_1}{1 - x_1} + \frac{x_2}{1 - x_2} + \dots + \frac{x_{100}}{1 - x_{100}} = 1.\]Find
\[\frac{x_1^2}{1 - x_1} + \frac{x_2^2}{1 - x_2} + \dots + \frac{x_{100}^2}{1 - x_{100}}.\].### Response: I notice that the given expression is similar to the second equation, except that the numerators are squared.
I wonder if I can use some algebraic manipulation to relate the two equations.
I recall that $\frac{x^2}{1 - x} = \frac{x}{1 - x} + \frac{x^2 - x}{1 - x} = \frac{x}{1 - x} + x$.
I apply this identity to each term in the expression and simplify.
I get \[\frac{x_1^2}{1 - x_1} + \frac{x_2^2}{1 - x_2} + \dots + \frac{x_{100}^2}{1 - x_{100}} = \left(\frac{x_1}{1 - x_1} + \frac{x_2}{1 - x_2} + \dots + \frac{x_{100}}{1 - x_{100}}\right) + \left(x_1 + x_2 + \dots + x_{100}\right).\]
I substitute the given values of the expressions in parentheses and simplify further.
I get \[\frac{x_1^2}{1 - x_1} + \frac{x_2^2}{1 - x_2} + \dots + \frac{x_{100}^2}{1 - x_{100}} = 1 + 1 = 2.\]