Adding and Subtracting Rational Expressions: Practice

Combine the top over the LCD $3x$.

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

The main bar means division, so multiply by the reciprocal of $x-3$.

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

The key move is $3 - x = -(x-3)$, which exposes the common factor. The restrictions are $x \neq 0$ and $x \neq 3$.

Factor both denominators: $x^2 - 9 = (x-3)(x+3)$ and $x^2 + 6x + 9 = (x+3)^2$. Taking each factor to its highest power gives the LCD $(x-3)(x+3)^2$.

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

Expand and collect the numerator: $3x + 9 + 2x - 6$.

$\frac{5x+3}{(x-3)(x+3)^2}$

The LCD is $(x-2)(x+2)$. Build both fractions up and bracket the entire numerator being subtracted.

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

Expand each product: $(2x+1)(x+2) = 2x^2 + 5x + 2$ and $(x+3)(x-2) = x^2 + x - 6$. Now distribute the minus sign across all three terms of the second product.

$\frac{2x^2 + 5x + 2 - x^2 - x + 6}{(x-2)(x+2)} = \frac{x^2 + 4x + 8}{(x-2)(x+2)}$

Stopping the minus sign after the $x^2$ would give $x^2 + 6x - 4$, the classic wrong answer.

Factor $x^2 - 1 = (x-1)(x+1)$, which is already the LCD for all three fractions. Build the first two up and bracket the numerator being subtracted.

$\frac{(x+1) - 2(x-1) + 3}{x^2-1}$

Distribute the minus across both terms of $2(x-1) = 2x - 2$.

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

Failing to distribute would give $x + 1 - 2x - 2 + 3 = 2 - x$, which is wrong.

Factor all three denominators.

$x^2-x-2 = (x-2)(x+1), \qquad x^2-4x+4 = (x-2)^2, \qquad x^2-4 = (x-2)(x+2)$

The distinct factors are $x-2$, $x+1$, and $x+2$. The highest power of $x-2$ in any single denominator is $2$, and the others reach only the first power.

$\text{LCD} = (x-2)^2(x+1)(x+2)$

Multiplying all three denominators would give $(x-2)^4(x+1)(x+2)$, which carries two surplus copies of $x-2$.

Factor the first denominator: $x^2 + 2x - 3 = (x+3)(x-1)$, which is the LCD. The second fraction is missing the factor $x+3$.

$\frac{(x+1) - (x+3)}{(x+3)(x-1)}$

Distribute the minus sign across both terms.

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

A check at $x=0$: the original is $\frac{1}{-3} - \frac{1}{-1} = \frac{2}{3}$, and the answer gives $\frac{-2}{-3} = \frac{2}{3}$.

Factor both denominators: $x^2 + x = x(x+1)$ and $x^2 - x = x(x-1)$. They share the factor $x$, so the LCD is $x(x+1)(x-1)$, not the product of the two.

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

Collect the numerator.

$\frac{5x-1}{x(x+1)(x-1)}$

Combine over the LCD $(x-2)(x+2)$, bracketing the numerator being subtracted.

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

The first two options are that answer written two ways, and the third is the same again, since $\frac{-4}{4-x^2} = \frac{-4}{-(x^2-4)} = \frac{4}{x^2-4}$. Only $\frac{2x}{x^2-4}$ is different: it is what you get by adding the two fractions instead of subtracting.

Combine the left side over the LCD $(x-1)(x+1) = x^2-1$.

$\frac{A(x+1) + B(x-1)}{x^2-1} = \frac{(A+B)x + (A-B)}{x^2-1}$

The denominators already match, so the numerators must match term by term: $A + B = 5$ and $A - B = 1$. Adding those equations gives $2A = 6$.

$A = 3, \qquad B = 2$

Check: $3(x+1) + 2(x-1) = 5x + 1$.

Factor $x^2 - 1 = (x-1)(x+1)$, so the LCD of all three is $(x-1)(x+1)$. Build the first two up by $x-1$.

$\frac{x(x-1) - (x-1) - (x-1)}{(x-1)(x+1)}$

Every term carries a factor of $x-1$, so pull it out of the numerator.

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

The restrictions $x \neq 1$ and $x \neq -1$ both survive, even though $x-1$ cancelled.

Combine the top over $x$ and the bottom over $x^2$, factoring the bottom as a difference of squares.

$2 - \frac{1}{x} = \frac{2x-1}{x}, \qquad 4 - \frac{1}{x^2} = \frac{4x^2-1}{x^2} = \frac{(2x-1)(2x+1)}{x^2}$

Multiply by the reciprocal of the bottom and cancel the shared factor $2x-1$.

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

The restrictions are $x \neq 0$, from the $\frac{1}{x}$ terms, and $x \neq \pm\frac{1}{2}$, since the bottom of the big fraction is zero exactly when $4 = \frac{1}{x^2}$.

Factor $x^2 - 4 = (x-2)(x+2)$, which is the LCD. The second fraction is missing the factor $x-2$.

$\frac{(x+1) - (x-2)}{(x-2)(x+2)}$

Distribute the minus sign across both terms.

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

Stopping the minus after the $x$ would give $x + 1 - x - 2 = -1$, the most common wrong answer here.