Adding and Subtracting Rational Expressions: Practice

The denominators share no factor, so the LCD is $(x-3)(x+3) = x^2 - 9$. Build each fraction up to it.

$\frac{x+3}{(x-3)(x+3)} + \frac{x-3}{(x-3)(x+3)} = \frac{(x+3) + (x-3)}{x^2-9}$

Collect the numerator.

$\frac{2x}{x^2-9}$

Had the problem been a subtraction, the numerator would have been $(x+3) - (x-3) = 6$ instead.

Factor both denominators completely.

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

The distinct factors are $x-2$ and $x+2$. The highest power of $x-2$ in a single denominator is $1$; the highest power of $x+2$ is $2$.

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

The product of the two denominators is $(x-2)(x+2)^3$, which carries one $x+2$ too many.

Factor: $x^2 - x = x(x-1)$, so the LCD is $x(x-1)$. The second fraction is missing a factor of $x$.

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

Collect the like terms in the numerator.

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

The numerator vanishes only at $x = 2$, which is not a zero of the denominator, so nothing cancels.

The denominators match, so subtract the numerators over the single denominator.

$\frac{x^2 - 9}{x-3}$

The numerator is a difference of squares, so factor it and cancel.

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

The answer is $x+3$, and it still carries the restriction $x \neq 3$ from the original denominator.

Factor the first denominator: two numbers multiplying to $-6$ and adding to $-1$ are $-3$ and $2$.

$x^2 - x - 6 = (x-3)(x+2)$

So the denominators are $(x-3)(x+2)$ and $x+2$. Setting each factor to zero gives $x = 3$ and $x = -2$, and those are the only excluded values.

The denominators share no factor, so the LCD is $(x+1)(x-1) = x^2 - 1$. Build each fraction up by the factor it is missing.

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

The $-x$ and $+x$ cancel.

$\frac{x^2+1}{x^2-1}$

The numerator $x^2+1$ does not factor over the reals, so nothing cancels.

The denominators look different but they are negatives of each other, since $4 - x = -(x-4)$. Rewrite the second fraction over $x-4$.

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

Now the denominators match, so combine the numerators.

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

The answer is $-1$, with $x \neq 4$. Treating $4-x$ as if it were $x-4$ gives $\frac{x+4}{x-4}$, which is wrong.

Restrictions come from the original denominators, so factor those first: $x - 2$ and $x^2 - 2x = x(x-2)$. They vanish at $x = 2$ and at $x = 0$, so both values are excluded before any algebra happens.

$\frac{5}{x-2} - \frac{10}{x(x-2)} = \frac{5x - 10}{x(x-2)} = \frac{5(x-2)}{x(x-2)} = \frac{5}{x}$

The factor $x-2$ cancelled, but cancelling never returns a value to the domain. The simplified form hides $x \neq 2$, so you must carry it.

The distinct factor is $x$, and the highest power it reaches in a single denominator is $2$, so the LCD is $x^2$ (not $x^3$).

$\frac{1}{x} \cdot \frac{x}{x} = \frac{x}{x^2}$

Now add over the common denominator.

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

Numeric coefficients follow the same rule as the letters: take the least common multiple of $2$ and $3$, and one copy of $x$. The LCD is $6x$.

$\frac{3}{2x} = \frac{9}{6x}, \qquad \frac{5}{3x} = \frac{10}{6x}$

Now the pieces are the same size.

$\frac{9}{6x} + \frac{10}{6x} = \frac{19}{6x}$

Factor the first denominator: $x^2 - x - 2 = (x-2)(x+1)$. That is the LCD, and the second fraction is missing $x+1$.

$\frac{5}{(x-2)(x+1)} - \frac{2(x+1)}{(x-2)(x+1)} = \frac{5 - (2x+2)}{(x-2)(x+1)}$

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

$\frac{5 - 2x - 2}{(x-2)(x+1)} = \frac{3-2x}{(x-2)(x+1)}$

Forgetting to distribute would give $5 - 2x + 2 = 7 - 2x$, which is wrong.

Factor both denominators.

$x^2 - x - 6 = (x-3)(x+2), \qquad x^2 - 9 = (x-3)(x+3)$

They share the factor $x-3$, which therefore appears just once in the LCD, alongside the two factors they do not share.

$\text{LCD} = (x-3)(x+2)(x+3)$

Multiplying the denominators would repeat $x-3$ and force a cancellation later.