Algebraic Fractions: Practice

Factor top and bottom: $x^2 + 6x = x(x + 6)$ and $2x + 12 = 2(x + 6)$.

$\frac{x^2 + 6x}{2x + 12} = \frac{x(x + 6)}{2(x + 6)} = \frac{x}{2}$

The greatest common factor of the numerator is $4$.

$\frac{8x + 12}{4} = \frac{4(2x + 3)}{4} = 2x + 3$

Cancel the shared $x + 3$, then reduce $\tfrac{x^2}{x}$.

$\frac{x + 3}{x} \cdot \frac{x^2}{x + 3} = \frac{x^2(x + 3)}{x(x + 3)} = \frac{x^2}{x} = x$

Factor $2x + 4 = 2(x + 2)$, then cancel the shared $x + 2$ and the $x$.

$\frac{2x + 4}{x} \cdot \frac{3x}{x + 2} = \frac{2(x + 2) \cdot 3x}{x(x + 2)} = \frac{6x}{x} = 6$

The original is undefined at $x = 0$ and $x = -2$, so this holds for $x \neq 0,\; x \neq -2$.

Factor $x^2 + 2x = x(x + 2)$, then multiply by the reciprocal $\tfrac{3}{x + 2}$.

$\frac{x(x + 2)}{6} \cdot \frac{3}{x + 2} = \frac{3x(x + 2)}{6(x + 2)} = \frac{3x}{6} = \frac{x}{2}$

Multiply by the reciprocal $\tfrac{x + 1}{5x}$; the $x + 1$ cancels.

$\frac{10x^2}{x + 1} \cdot \frac{x + 1}{5x} = \frac{10x^2}{5x} = 2x$

The LCD of $4$ and $3$ is $12$.

$\frac{x}{4} + \frac{2x}{3} = \frac{3x}{12} + \frac{8x}{12} = \frac{11x}{12}$

The LCD of $x$ and $2x$ is $2x$; rewrite $\tfrac{4}{x}$ as $\tfrac{8}{2x}$.

$\frac{4}{x} + \frac{3}{2x} = \frac{8}{2x} + \frac{3}{2x} = \frac{11}{2x}$

The LCD of $2$ and $5$ is $10$.

$\frac{x}{2} - \frac{x}{5} = \frac{5x}{10} - \frac{2x}{10} = \frac{3x}{10}$

Factor the numerator as $3(x - 4)$, then cancel the shared $x - 4$.

$\frac{3x - 12}{x - 4} = \frac{3(x - 4)}{x - 4} = 3$

The original is undefined at $x = 4$, so this holds for $x \neq 4$.

Add the numerators over the common denominator, then simplify.

$\frac{x + 1}{x} + \frac{2x - 1}{x} = \frac{3x}{x} = 3$

The constants $+1$ and $-1$ cancel, leaving $3x$ over $x$. The original is undefined at $x = 0$, so this holds for $x \neq 0$.

The LCD of $x$ and $2$ is $2x$; rewrite each fraction and subtract.

$\frac{5}{x} - \frac{1}{2} = \frac{10}{2x} - \frac{x}{2x} = \frac{10 - x}{2x}$