Adding and Subtracting Rational Expressions: Additional Practice (Set 1)

Combine the top and the bottom, each over the denominator $x+1$.

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

Divide by multiplying by the reciprocal; the $x+1$ factors cancel.

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

The minus comes from $1 - (x+1) = -x$, so the sign is decided in the bottom. The restrictions are $x \neq -1$ and $x \neq 0$.

The LCD is $(x+2)(x-1)$, so each numerator is multiplied by the factor its fraction is missing.

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

Expand both squares: $(x^2 - 2x + 1) + (x^2 + 4x + 4)$.

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

The numerator has no real roots, so it shares no factor with the denominator and the fraction is in lowest terms.

Factor all three denominators.

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

The distinct factors are $x-3$ and $x+3$. The highest power of $x-3$ in a single denominator is $2$, and the highest power of $x+3$ is $1$.

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

The choice $(x-3)(x+3)$ is not a common denominator at all, because $(x-3)^2$ does not divide it. Multiplying all three denominators together would give $(x-3)^3(x+3)^2$, a common denominator that carries two surplus factors.

Factor both denominators.

$x^2 - 2x - 8 = (x-4)(x+2), \qquad x^2 - x - 12 = (x-4)(x+3)$

Set every factor to zero. The shared factor $x-4$ contributes $x=4$ once, and the other two contribute $x=-2$ and $x=-3$.

$x = 4, \quad x = -2, \quad x = -3$

Factor both denominators completely, starting by pulling out the common $x$.

$x^3 - x = x(x^2-1) = x(x-1)(x+1), \qquad x^2 + x = x(x+1)$

Every factor of the second denominator already appears in the first, so the first denominator is itself the LCD.

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

The product of the two denominators, $x^2(x-1)(x+1)^2$, carries a surplus $x$ and a surplus $x+1$.

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

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

Distribute the minus sign across both terms of $x-1$.

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

Failing to distribute gives $x - x - 1 + 1 = 0$, a wrong answer that looks suspiciously tidy.

Factor the right-hand denominator: $x^2 + x - 6 = (x-2)(x+3)$, which is exactly the LCD on the left. Combining the left side gives

$\frac{A(x+3) + B(x-2)}{(x-2)(x+3)} = \frac{x+13}{(x-2)(x+3)},$

so the numerators agree at every allowed $x$. That makes them equal as polynomials: their difference has degree at most $1$ and vanishes at infinitely many values, and only the zero polynomial can do that. A polynomial identity holds at every $x$ with no exceptions, including the two values the original fractions bar, so you may substitute those values freely even though the fractions themselves are undefined there.

Setting $x=2$ kills the $B$ term and gives $5A = 15$; setting $x=-3$ kills the $A$ term and gives $-5B = 10$.

$A = 3, \qquad B = -2$

Check: $3(x+3) - 2(x-2) = 3x + 9 - 2x + 4 = x + 13$.

Combine the top over the LCD $x(x+1)$, bracketing the fraction being subtracted.

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

The top and the bottom are now the same fraction, so multiplying by the reciprocal of the bottom gives

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

The restrictions are $x \neq 0$ and $x \neq -1$, which the answer $1$ shows no trace of.

Factor $x^2 - 9 = (x-3)(x+3)$, which is the LCD. The first fraction is missing the factor $x-3$, not $x+3$.

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

The numerator $x^2 - 3x + 9$ has discriminant $9 - 36 < 0$, so it does not factor over the reals and nothing cancels.

The minus sign in front of $2(x+1)$ applies to the whole product, so both of its terms change sign.

$4(x-3) - 2(x+1) = 4x - 12 - 2x - 2$

Collect the like terms.

$2x - 14$

The student turned $-2$ into $+2$, which is the single most common error in this lesson. Bracketing the numerator being subtracted before expanding prevents it.

Factor: $x^2 - 4 = (x-2)(x+2)$ and $x^2 - 4x + 4 = (x-2)^2$. Taking each factor to its highest power gives the LCD $(x-2)^2(x+2)$.

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

Distribute the minus sign across both terms.

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

A check at $x=0$: the original is $\frac{1}{-4} - \frac{1}{4} = -\frac{1}{2}$, and the answer gives $\frac{-4}{4 \cdot 2} = -\frac{1}{2}$.

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

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

Collect the numerator.

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

Adding the tops and the bottoms separately would give $\frac{2}{2x+2}$, which is not even close: at $x=2$ the true rate is $\frac{6}{8}$ of a tank per hour, while that wrong form gives $\frac{1}{3}$.