Algebraic Fractions: Additional Practice (Set 2)

The greatest common factor of $12x^3$ and $18x^2$ is $6x^2$.

$\frac{12x^3}{18x^2} = \frac{6x^2 \cdot 2x}{6x^2 \cdot 3} = \frac{2x}{3}$

Multiply across, then cancel: $\tfrac{9}{6} \cdot \tfrac{5}{10} = \tfrac{3}{4}$, and one $x$ cancels.

$\frac{5x^2}{6} \cdot \frac{9}{10x} = \frac{45x^2}{60x} = \frac{3x}{4}$

Distribute the minus sign across both terms of $2x - 5$.

$\frac{6x + 1}{x + 3} - \frac{2x - 5}{x + 3} = \frac{6x + 1 - 2x + 5}{x + 3} = \frac{4x + 6}{x + 3}$

The $-5$ becomes $+5$.

Multiply by the reciprocal $\tfrac{x + 2}{3x}$; the $x + 2$ and the $x$ both cancel.

$\frac{x}{x + 2} \cdot \frac{x + 2}{3x} = \frac{x(x + 2)}{3x(x + 2)} = \frac{1}{3}$

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

The LCD is $2x$; rewrite $\tfrac{1}{x}$ as $\tfrac{2}{2x}$.

$\frac{5}{2x} + \frac{1}{x} = \frac{5}{2x} + \frac{2}{2x} = \frac{7}{2x}$

The factor $2x$ on top and $6x$ on the bottom share $2x$, leaving $\tfrac{1}{3}$.

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

The shared factor $3x$ cancels, and the numerator keeps its negative sign.

$\frac{-6x^2}{9x} = \frac{-3x \cdot 2x}{3x \cdot 3} = -\frac{2x}{3}$

Factor $3x + 18 = 3(x + 6)$, cancel the shared $x + 6$, and reduce.

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

The original is undefined at $x = -6$, so this holds for $x \neq -6$.

Both the $x$ and the $x + 1$ appear once on top and once on the bottom.

$\frac{x}{x + 1} \cdot \frac{x + 1}{x} = \frac{x(x + 1)}{(x + 1)x} = 1$

The original is undefined at $x = 0$ and $x = -1$, so this holds for $x \neq 0,\; x \neq -1$.

Multiply by the reciprocal $\tfrac{x - 2}{3x}$; the $x - 2$ cancels.

$\frac{9x}{x - 2} \cdot \frac{x - 2}{3x} = \frac{9x}{3x} = 3$

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

Cancel the shared $x + 5$, then reduce $\tfrac{8x}{4x}$.

$\frac{x + 5}{4x} \cdot \frac{8x}{x + 5} = \frac{8x(x + 5)}{4x(x + 5)} = \frac{8x}{4x} = 2$

The original is undefined at $x = 0$ and $x = -5$, so this holds for $x \neq 0,\; x \neq -5$.

Factor $6x + 18 = 6(x + 3)$, then multiply by the reciprocal $\tfrac{10}{x + 3}$.

$\frac{6(x + 3)}{5} \cdot \frac{10}{x + 3} = \frac{60(x + 3)}{5(x + 3)} = \frac{60}{5} = 12$

The original divisor is zero at $x = -3$, so this holds for $x \neq -3$.