Adding and Subtracting Rational Expressions: Additional Practice (Set 2)

The three denominators share no factors, so the LCD is their product, $x(x+1)(x+2)$. Each fraction is missing the two factors it does not already have.

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

Expand each product in the numerator: $(x^2+3x+2) + (x^2+2x) + (x^2+x)$.

$\frac{3x^2 + 6x + 2}{x(x+1)(x+2)}$

As a check, at $x=1$ the original is $1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}$, and the answer gives $\frac{3+6+2}{1 \cdot 2 \cdot 3} = \frac{11}{6}$.

Factor both denominators: $x^2 + 3x + 2 = (x+1)(x+2)$ and $x^2 - 1 = (x-1)(x+1)$. They share $x+1$, so the LCD is $(x+1)(x+2)(x-1)$.

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

Distribute the minus sign across both terms of $3(x+2) = 3x + 6$.

$\frac{5x - 5 - 3x - 6}{(x+1)(x+2)(x-1)} = \frac{2x-11}{(x+1)(x+2)(x-1)}$

Stopping the minus after the $3x$ would give $2x + 1$, which is wrong.

Combine the top over $x$ and the bottom over $x^2$.

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

Divide by multiplying by the reciprocal, then cancel the shared factors $x$ and $x+1$.

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

The restrictions are $x \neq 0$ and $x \neq \pm 1$.

The LCD is $(x-3)(x+3) = x^2 - 9$. Build the last two fractions up to it, bracketing the numerator being subtracted.

$\frac{2x - (x+3) + (x-3)}{x^2-9} = \frac{2x - x - 3 + x - 3}{x^2-9} = \frac{2x-6}{x^2-9}$

Factor the numerator and cancel.

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

The restrictions $x \neq 3$ and $x \neq -3$ both remain, even though $x-3$ cancelled.

The two denominators are negatives of each other, since $3 - x = -(x-3)$. Rewrite the second fraction over $x-3$.

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

Now the denominators match and the numerators cancel.

$\frac{1}{x-3} + \frac{-1}{x-3} = \frac{1-1}{x-3} = 0$

The expression is $0$ everywhere it is defined, which is everywhere except $x = 3$.

Factor $x^2 - 16 = (x-4)(x+4)$, which is the LCD. Build the last two fractions up and bracket the numerator being subtracted.

$\frac{4 + (x+4) - (x-4)}{x^2-16}$

Distribute the minus sign across both terms of $x-4$.

$\frac{4 + x + 4 - x + 4}{x^2-16} = \frac{12}{x^2-16}$

Both of the tempting wrong numerators are on offer here. Losing the sign on the $-4$ gives $4 + x + 4 - x - 4 = 4$, and adding the third fraction instead of subtracting it gives $4 + x + 4 + x - 4 = 2x + 4$. Distributing the minus sign deliberately is what separates the three.

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

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

Distribute the minus sign.

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

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

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

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

Distribute the minus sign.

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

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

The LCD is $(x-2)(x+2) = x^2 - 4$. Build the first two fractions up to it and bracket the numerator being subtracted.

$\frac{(x-2) + (x+2) - 2x}{x^2-4} = \frac{2x - 2x}{x^2-4} = 0$

The expression is identically zero, but only where it exists. The original denominators vanish at $x = 2$ and $x = -2$, so those values were never in the domain and remain excluded.

The denominators $2-x$ and $x-2$ are negatives of each other, so multiply the numerator and the denominator by $-1$, which leaves the value unchanged.

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

Flipping the denominator's sign flips the numerator's sign with it. Rewriting $2-x$ as $x-2$ and leaving the $3$ alone would change the value of the expression.

Combine the top and the bottom over the same LCD, $xy$.

$\frac{1}{x} + \frac{1}{y} = \frac{y+x}{xy}, \qquad \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy}$

Divide by multiplying by the reciprocal; the $xy$ factors cancel.

$\frac{x+y}{xy} \cdot \frac{xy}{y-x} = \frac{x+y}{y-x}$

The order in the bottom matters: it is $y - x$, not $x - y$, because $\frac{1}{x} - \frac{1}{y}$ puts $y$ first in the numerator.

There is one distinct factor, $x-2$, and its highest power in a single denominator is $3$, so the LCD is $(x-2)^3$. (The product of the three denominators would be $(x-2)^6$, three copies too many.)

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

Expand and collect: $x^2 - 4x + 4 + x - 2 + 1$.

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