Algebraic Fractions: Practice

Factor the numerator as $5(x + 3)$, then cancel the shared $x + 3$.

$\frac{5x + 15}{x + 3} = \frac{5(x + 3)}{x + 3} = 5$

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

Factor the denominator as $2(x + 2)$, then cancel the shared $x + 2$.

$\frac{x + 2}{2x + 4} = \frac{x + 2}{2(x + 2)} = \frac{1}{2}$

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

Cancel the shared $x + 1$ and reduce $\tfrac{4}{2}$.

$\frac{x + 1}{2} \cdot \frac{4}{x + 1} = \frac{4(x + 1)}{2(x + 1)} = \frac{4}{2} = 2$

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

Cancel the shared $x + 2$, then reduce $\tfrac{3x}{6}$.

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

Multiply by the reciprocal $\tfrac{8}{x^2}$, then cancel.

$\frac{x}{4} \div \frac{x^2}{8} = \frac{x}{4} \cdot \frac{8}{x^2} = \frac{8x}{4x^2} = \frac{2}{x}$

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

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

The original is undefined at $x = 0$ and $x = -3$, so this holds for $x \neq 0,\; x \neq -3$.

Dividing by $3$ is multiplying by $\tfrac{1}{3}$.

$\frac{9}{x} \div 3 = \frac{9}{x} \cdot \frac{1}{3} = \frac{9}{3x} = \frac{3}{x}$

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

$\frac{2}{x} + \frac{3}{x^2} = \frac{2x}{x^2} + \frac{3}{x^2} = \frac{2x + 3}{x^2}$

The LCD of $x$ and $2$ is $2x$.

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

The LCD is $2x$; rewrite $\tfrac{3}{x}$ as $\tfrac{6}{2x}$.

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

Distribute the minus sign to every term of $2x - 4$.

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

The $-4$ becomes $+4$, giving $2 + 4 = 6$.

Set the denominator equal to zero and solve.

$2x - 6 = 0 \;\Rightarrow\; x = 3$

So $x = 3$ is excluded.