Question #N740

If $x \neq 0$, what is the value of the expression $\frac{1}{x} + \frac{1}{x+1}$ expressed as a single fraction?
A. $\frac{1}{x(x+1)}$
B. $\frac{2}{x(x+1)}$
C. $\frac{2x+1}{x(x+1)}$
D. $\frac{x+2}{x(x+1)}$

Correct Answer is: C

To add fractions, they must have a common denominator. The least common denominator for $x$ and $x+1$ is $x(x+1)$. Multiplying the first fraction by $\frac{x+1}{x+1}$ and the second fraction by $\frac{x}{x}$, we get: $\frac{1}{x} + \frac{1}{x+1} = \frac{1}{x} \cdot \frac{x+1}{x+1} + \frac{1}{x+1} \cdot \frac{x}{x} = \frac{x+1}{x(x+1)} + \frac{x}{x(x+1)}$. Adding the numerators gives us $\frac{x+1 + x}{x(x+1)} = \frac{2x+1}{x(x+1)}$.