Rational expressions — simplification, operations: Worked examples

The scenario matrix

Before any example, let's map the whole territory. Each row is a case class — a distinct kind of thing that can happen. Every worked example below is tagged with the cell it covers.

#	Case class	What is special about it	Covered by
A	Clean simplify, common factor cancels	A factor on top matches one on bottom	Ex 1
B	The "hole" trap	A cancelled factor leaves a hidden restriction	Ex 2
C	Sign flip: $(a - b)$ vs $(b - a)$	Factors that are negatives of each other	Ex 3
D	Multiply / divide with cross-cancel	Two fractions, factors cancel across them	Ex 4
E	Add/subtract, unlike denominators, LCD	Build a common denominator first	Ex 5
F	Degenerate / zero input	Numerator becomes $0$ ; result is legitimately $0$	Ex 6
G	Limiting behaviour ( $x \to \infty$ )	What the expression approaches for huge $x$	Ex 7
H	Real-world word problem (rates)	Meaning + units, not just algebra	Ex 8
I	Exam twist: complex fraction	A fraction stacked inside a fraction	Ex 9

We will need the picture of a hole more than once, so let's draw it first.

Figure — Rational expressions — simplification, operations

Forecast: guess the final fraction before reading on. (Hint: both parts factor.)

Factor the top. Why this step? Cancelling only works on multiplied factors, so we must expose them. $x^{2} - 4 = (x - 2) (x + 2) (difference of squares a^{2} - b^{2} = (a - b) (a + b))$
Factor the bottom. Why? Same reason — find factors, not terms. Two numbers multiplying to $6$ , adding to $5$ : that's $2$ and $3$ . $x^{2} + 5 x + 6 = (x + 2) (x + 3)$
List restrictions from the ORIGINAL bottom. Why now? The forbidden values are decided before we cancel anything. $x + 2 \neq = 0 \Rightarrow x \neq = - 2, x + 3 \neq = 0 \Rightarrow x \neq = - 3$
Cancel the shared factor $(x + 2)$ . Why allowed? $\frac{a \cdot b}{a \cdot c} = \frac{b}{c}$ when $a \neq = 0$ . $\frac{( x - 2 ) ( x + 2 )}{( x + 2 ) ( x + 3 )} = \frac{x - 2}{x + 3}$

Answer: $\frac{x - 2}{x + 3}$ , where $x \neq = - 2, - 3$ .

Verify: put $x = 0$ : original $= \frac{- 4}{6} = - \frac{2}{3}$ ; simplified $= \frac{- 2}{3} = - \frac{2}{3}$ . ✓

Forecast: the tidy form is obviously $\frac{1}{x + 2}$ — but how many forbidden values?

Spot restrictions first. Why? Both bottom factors matter before we simplify. $x - 3 \neq = 0 \Rightarrow x \neq = 3, x + 2 \neq = 0 \Rightarrow x \neq = - 2$
Cancel $(x - 3)$ . Why? Common factor top and bottom. $\frac{( x - 3 )}{( x - 3 ) ( x + 2 )} = \frac{1}{x + 2}$
Carry BOTH restrictions forward. Why? The cancelled $(x - 3)$ leaves a hole — look at s01: the curve $\frac{1}{x + 2}$ is drawn, but there's an open circle at $x = 3$ .

Answer: $\frac{1}{x + 2}$ , where $x \neq = 3, - 2$ .

Verify: at $x = 3$ the original is $\frac{0}{0}$ (undefined) while $\frac{1}{x + 2} = \frac{1}{5}$ — different, confirming $x = 3$ must stay excluded.

Forecast: $x - 3$ and $3 - x$ look almost the same — do they cancel?

Factor the bottom. Why? Expose factors. $x^{2} - 9 = (x - 3) (x + 3)$ .
Rewrite the top so factors match. Why this step? $3 - x$ and $x - 3$ are negatives of each other: $3 - x = - (x - 3)$ . Pulling out $- 1$ makes the shared factor visible. $\frac{3 - x}{( x - 3 ) ( x + 3 )} = \frac{- ( x - 3 )}{( x - 3 ) ( x + 3 )}$
Restrictions from original bottom. $x \neq = 3, x \neq = - 3$ .
Cancel $(x - 3)$ , keep the minus. Why keep it? The $- 1$ is a genuine factor, not garbage to drop. $\frac{- ( x - 3 )}{( x - 3 ) ( x + 3 )} = \frac{- 1}{x + 3}$

Answer: $\frac{- 1}{x + 3}$ , where $x \neq = 3, - 3$ .

Verify: $x = 0$ : original $= \frac{3}{- 9} = - \frac{1}{3}$ ; answer $= \frac{- 1}{3} = - \frac{1}{3}$ . ✓

Forecast: division flips the second fraction — then watch factors vanish across the whole product.

Flip the divisor (multiply by reciprocal). Why? Dividing by $\frac{c}{d}$ equals multiplying by $\frac{d}{c}$ . $\frac{x ^{2} + 2 x}{x + 3} \cdot \frac{x ^{2} - 9}{x}$
Factor everything. Why first? So factors can cancel across both fractions. $\frac{x ( x + 2 )}{x + 3} \cdot \frac{( x - 3 ) ( x + 3 )}{x}$
Collect all restrictions. Why every denominator, including the flipped one? Each zero was forbidden in the original setup: $x \neq = - 3, x \neq = 0$ , and from the original divisor top $x \neq = 3$ .
Cancel $x$ and $(x + 3)$ . $\frac{x ( x + 2 ) ( x + 3 ) ( x - 3 )}{( x + 3 ) x} = (x + 2) (x - 3) = x^{2} - x - 6$

Answer: $x^{2} - x - 6$ , where $x \neq = - 3, 0, 3$ .

Verify: $x = 1$ : first fraction $\frac{1 + 2}{4} = \frac{3}{4}$ ; divisor $\frac{1}{1 - 9} = - \frac{1}{8}$ ; $\frac{3}{4} \div (- \frac{1}{8}) = - 6$ . Formula: $1 - 1 - 6 = - 6$ . ✓

Forecast: you can't subtract until both bottoms match. What's the smallest common denominator?

Factor the first denominator. $x^{2} - 4 = (x - 2) (x + 2)$ .
Choose the LCD. Why $(x - 2) (x + 2)$ ? It contains every factor present, each to its highest power. The second fraction is missing $(x - 2)$ .
Rebuild the second fraction. Why multiply by $\frac{x - 2}{x - 2}$ ? That equals $1$ , so the value is unchanged. $\frac{x}{( x - 2 ) ( x + 2 )} - \frac{2 ( x - 2 )}{( x - 2 ) ( x + 2 )}$
Combine numerators — mind the minus sign over the whole bracket. Why brackets? The subtraction distributes to both terms of $2 (x - 2)$ . $\frac{x - 2 ( x - 2 )}{( x - 2 ) ( x + 2 )} = \frac{x - 2 x + 4}{( x - 2 ) ( x + 2 )} = \frac{4 - x}{( x - 2 ) ( x + 2 )}$

Answer: $\frac{4 - x}{( x - 2 ) ( x + 2 )}$ , where $x \neq = 2, - 2$ .

Verify: $x = 0$ : original $\frac{0}{- 4} - \frac{2}{2} = 0 - 1 = - 1$ ; answer $\frac{4}{( - 2 ) ( 2 )} = \frac{4}{- 4} = - 1$ . ✓

Forecast: one input makes the top zero — is the answer $0$ or undefined?

Factor the top. Why? To see when it hits zero. $x^{2} - 5 x = x (x - 5)$ .
At $x = 5$ : top $= 5 (0) = 0$ , bottom $= 9 \neq = 0$ . Why this matters: $\frac{0}{nonzero} = 0$ — perfectly legal, the expression equals $0$ . $R (5) = \frac{0}{9} = 0$
At $x = 0$ : top $= 0$ , bottom $= 4 \neq = 0$ , so again $R (0) = \frac{0}{4} = 0$ . Why check separately? $x = 0$ is a root of the top, not of the bottom, so it is allowed.
Contrast with the forbidden input. Only $x = - 4$ is excluded (makes bottom zero).

Answer: $R (5) = 0, R (0) = 0$ ; the sole restriction is $x \neq = - 4$ .

Verify: direct substitution $\frac{25 - 25}{9} = 0$ and $\frac{0 - 0}{4} = 0$ . ✓

Forecast: guess — does it blow up, die to zero, or settle on a number?

Divide top and bottom by the highest power $x^{2}$ . Why this tool? For huge $x$ , comparing everything to the biggest term reveals what dominates. $\frac{3 x ^{2} + 2 x}{x ^{2} - 7} = \frac{3 + \frac{2}{x}}{1 - \frac{7}{x ^{2}}}$
See what tiny pieces do. Why? As $x$ grows, $\frac{2}{x} \to 0$ and $\frac{7}{x ^{2}} \to 0$ . This is the idea of a limit — the value the expression approaches.
Read off the survivor. $\to \frac{3 + 0}{1 - 0} = 3$

The curve flattens toward the height $3$ — a horizontal asymptote (see s02).

Answer: the expression approaches $3$ as $x \to \infty$ (and as $x \to - \infty$ ).

Verify: $x = 1000$ : $\frac{3 000 000 + 2000}{1 000 000 - 7} \approx 3.00199 \approx 3$ . ✓

Forecast: it's a subtraction of two fractions — units will be hours.

Write the difference. Why subtract? "Time saved" = slower time minus faster time. $t_{A} - t_{B} = \frac{6}{x} - \frac{6}{x + 2}$
LCD $= x (x + 2)$ ; rebuild each fraction. Why? Cannot subtract unlike denominators. $\frac{6 ( x + 2 )}{x ( x + 2 )} - \frac{6 x}{x ( x + 2 )}$
Combine numerators. $\frac{6 x + 12 - 6 x}{x ( x + 2 )} = \frac{12}{x ( x + 2 )}$
State restriction & meaning. $x \neq = 0, - 2$ (a rate of $0$ makes no physical sense anyway). Units: $\frac{length}{( rate ) ^{2}} =$ hours.

Answer: time saved $= \frac{12}{x ( x + 2 )}$ hours, $x \neq = 0, - 2$ .

Verify: $x = 4$ : $t_{A} = \frac{6}{4} = 1.5$ , $t_{B} = \frac{6}{6} = 1$ , saved $= 0.5$ h; formula $\frac{12}{4 \cdot 6} = \frac{12}{24} = 0.5$ . ✓

Forecast: a complex fraction — a fraction inside a fraction. Clear the little ones first.

Combine the top into one fraction. Why? Two stacked layers are hard to cancel; make the top a single fraction. LCD of $x$ and $2$ is $2 x$ . $\frac{1}{x} - \frac{1}{2} = \frac{2}{2 x} - \frac{x}{2 x} = \frac{2 - x}{2 x}$
Rewrite the whole thing as a division. Why? A fraction bar means "top divided by bottom." $\frac{\frac{2 - x}{2 x}}{x - 2} = \frac{2 - x}{2 x} \div (x - 2) = \frac{2 - x}{2 x} \cdot \frac{1}{x - 2}$
Handle the sign-flip factor. Why? $2 - x = - (x - 2)$ , so it cancels with $(x - 2)$ leaving a minus. $\frac{- ( x - 2 )}{2 x ( x - 2 )} = \frac{- 1}{2 x}$
Restrictions: original had $x \neq = 0$ (inside) and $x \neq = 2$ (outer bottom).

Answer: $\frac{- 1}{2 x}$ , where $x \neq = 0, 2$ .

Verify: $x = 4$ : original $\frac{\frac{1}{4} - \frac{1}{2}}{4 - 2} = \frac{- \frac{1}{4}}{2} = - \frac{1}{8}$ ; answer $\frac{- 1}{8} = - \frac{1}{8}$ . ✓

Recall check

Recall Why do cancelled factors still restrict the domain?

Because the original denominator was zero there — that value was never allowed, and cancelling only tidies the formula, it does not change what the expression "was."

Recall How do you handle

$(a - b)$ against $(b - a)$ ? Factor out $- 1$ : $(b - a) = - (a - b)$ . Then the factor cancels and you keep the leftover minus sign.

Which cell has a genuinely

0

answer (not undefined)?

Cell F — top is zero but bottom is nonzero, so

\frac{0}{nonzero} = 0

x \to \infty

\frac{3 x ^{2} + 2 x}{x ^{2} - 7}

approaches what?

3

(ratio of leading coefficients).

The value forbidden by a cancelled factor creates what on the graph?

A hole (removable discontinuity).

Related: Polynomial factoring · Domain and range · Complex fractions · Rational equations · Polynomial long division · Limits and continuity · Partial fraction decomposition