← mapAlgebra Core

Rational Expressions & Equations

⚗ Dr. Möbius, from the lab

Everything you learned about fractions in the Fractions and Rationals lesson applies here, word for word, symbol for symbol — because rational expressions ARE fractions. The numerator and denominator happen to be polynomials instead of integers, but the rules are the same theorems. The one new hazard: denominators that are zero are a genuine catastrophe, and you will track that domain landmine at every step.

THE BIG IDEA

A rational expression is a fraction with polynomial numerator and denominator; all fraction rules from Bedrock apply, but excluded values (zeros of the denominator) must be tracked throughout.

What a rational expression is

A rational expression is a ratio $\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials and $Q$ is not the zero polynomial. The domain is all real values of $x$ for which $Q(x) \ne 0$ . Division by zero is not a real number — it's an error in the universe — and we exclude those values at the outset.

Example: $\frac{x+2}{x^2-4}$ . The denominator is $x^2-4 = (x-2)(x+2)$ . It equals zero when $x = 2$ or $x = -2$ . Domain: all real $x$ except $x = 2$ and $x = -2$ .

Simplifying: factor, then cancel FACTORS

The Fractions lesson derived that $\frac{ac}{bc} = \frac{a}{b}$ for $c \ne 0$ — you cancel a common factor. For polynomials, you do the same thing, but you must factor first.

$\frac{x^2 - 4}{x^2 - 5x + 6} = \frac{(x+2)(x-2)}{(x-2)(x-3)} = \frac{x+2}{x-3}, \quad x \ne 2, x \ne 3.$

The cardinal sin: canceling terms, not factors. $\frac{x+2}{x+5} \ne \frac{2}{5}$ . The $x$ cannot be "canceled" because it's a term being added, not a factor being multiplied. You can only cancel something that multiplies the entire numerator against something that multiplies the entire denominator.

The restriction $x \ne 2$ survives even after simplification, because the original expression was not defined at $x = 2$ . The simplified form $\frac{x+2}{x-3}$ looks fine at $x=2$ , but it represents the same function as the original only on the common domain. Don't drop excluded values.

Multiplying and dividing

Multiply: same rule as numeric fractions — multiply numerators, multiply denominators, then simplify: $\frac{P_1}{Q_1} \cdot \frac{P_2}{Q_2} = \frac{P_1 P_2}{Q_1 Q_2}.$

But factor BEFORE multiplying — it's much easier to cancel before computing huge products.

Example: $\frac{x^2-1}{x+3} \cdot \frac{x^2+6x+9}{x^2-x}$ .

Factor everything: $\frac{(x+1)(x-1)}{x+3} \cdot \frac{(x+3)^2}{x(x-1)} = \frac{(x+1)(x-1)(x+3)^2}{x(x-1)(x+3)} = \frac{(x+1)(x+3)}{x}.$

Division: multiply by the reciprocal (same as numeric division, from the Fractions lesson): $\frac{P_1}{Q_1} \div \frac{P_2}{Q_2} = \frac{P_1}{Q_1} \cdot \frac{Q_2}{P_2} = \frac{P_1 Q_2}{Q_1 P_2}.$

Adding and subtracting: the LCD is back

Same rule as numeric fractions: find the LCD, rewrite each fraction, then add/subtract numerators. The LCD of two rational expressions is the LCM of their denominators (factored form essential).

Example: $\frac{3}{x+1} + \frac{2}{x-1}$ .

LCD $= (x+1)(x-1)$ .

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

The reason LCD works is the same theorem from the Fractions lesson: $\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$ when $bd \ne 0$ . This is distributivity operating inside fractions.

Solving rational equations: the LCD trick and extraneous solutions

To solve an equation involving rational expressions, multiply both sides by the LCD to clear all denominators. But here is the critical warning: multiplying by an expression that might be zero can introduce extraneous solutions.

Example: Solve $\frac{2}{x-1} = \frac{3}{x+2}$ .

LCD $= (x-1)(x+2)$ . Multiply both sides: $2(x+2) = 3(x-1) \implies 2x+4 = 3x-3 \implies x = 7.$

Check: $x = 7$ makes neither denominator zero ( $7-1=6 \ne 0$ ; $7+2=9 \ne 0$ ). Substitute: $\frac{2}{6} = \frac{1}{3}$ and $\frac{3}{9} = \frac{1}{3}$ . Equal. Confirmed.

Example with extraneous solution: Solve $\frac{x}{x-2} + 1 = \frac{2}{x-2}$ .

Multiply by $(x-2)$ : $x + (x-2) = 2 \implies 2x - 2 = 2 \implies x = 2.$

But $x = 2$ makes the denominator zero in the original equation. Extraneous solution — no solution. The equation has no answer. When you multiplied by $(x-2)$ and $x = 2$ , you multiplied by zero, which is an illegal operation (you can't conclude $a = b$ from $0 \cdot a = 0 \cdot b$ ).

Always check every solution in the original equation.

🔬 SPECIMENS (worked examples)

Worked example 1 — simplifying with excluded values▸

Simplify $\frac{x^2 + 3x - 10}{x^2 - 4}$ and state the excluded values.

Factor numerator: $x^2+3x-10 = (x+5)(x-2)$ . (Need $pq=-10$ , $p+q=3$ : pair $(5,-2)$ .)

Factor denominator: $x^2-4 = (x+2)(x-2)$ .

Cancel the common factor $(x-2)$ : $\frac{(x+5)(x-2)}{(x+2)(x-2)} = \frac{x+5}{x+2}.$

Excluded values: $x=2$ (makes original denominator zero) and $x=-2$ (makes both original and simplified denominator zero).

The simplified form $\frac{x+5}{x+2}$ is valid for all $x$ except $x = \pm 2$ .

Worked example 2 — adding rational expressions▸

Add: $\frac{x}{x^2-1} + \frac{2}{x-1}$ .

Factor: $x^2-1 = (x+1)(x-1)$ .

LCD $= (x+1)(x-1)$ . The second fraction needs an extra factor of $(x+1)$ :

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

Excluded values: $x = \pm 1$ (throughout — they make the original denominators zero).

Check with $x = 2$ : LHS $= \frac{2}{3} + \frac{2}{1} = \frac{2}{3}+2 = \frac{8}{3}$ . RHS $= \frac{3(2)+2}{4-1} = \frac{8}{3}$ . Confirmed.

Worked example 3 — extraneous solution, caught▸

Solve: $\frac{x+1}{x-3} = \frac{4}{x-3} + 1$ .

Identify excluded values first: $x = 3$ makes both denominators zero; it's excluded.

Multiply through by $(x-3)$ : $x+1 = 4 + (x-3) = x+1.$

This gives $x+1 = x+1$ , which is true for all $x$ — an identity. But we can't include $x = 3$ . So the solution set is: all real $x$ except $x = 3$ .

This is the dependent/identity case from the Linear Equations lesson, showing up inside rational expressions. The equation is satisfied by every real number in the domain. The answer is NOT "all reals" — it is all reals except 3.

☠ KNOWN HAZARDS

Canceling terms instead of factors. $\frac{x+3}{x+5}$ cannot be simplified — the $x$ is a term being added, not a factor. Only cancel things that MULTIPLY the entire top or bottom.
Dropping excluded values after simplification. Even after canceling $(x-2)/(x-2)$ , the value $x=2$ is still excluded from the domain — it wasn't defined in the original expression.
Forgetting to check for extraneous solutions. If your "solution" makes any denominator zero, it's extraneous — it doesn't satisfy the original equation. Always substitute back into the ORIGINAL, not a simplified version.
LCD error: using the product of denominators instead of the LCM. For $\frac{1}{x(x-1)}$ and $\frac{1}{(x-1)(x+2)}$ , the LCD is $x(x-1)(x+2)$ , not $x(x-1)^2(x+2)$ .

TL;DR

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A rational expression is a polynomial fraction. The domain excludes all values making the denominator zero — track these throughout.
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Simplify by factoring numerator and denominator, then canceling COMMON FACTORS (never terms). Excluded values persist.
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Multiply/divide: factor first, cancel, then multiply. Division flips the second fraction.
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Add/subtract: find the LCD (factor denominators), build equivalent fractions, then combine numerators.
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Solving: clear denominators with the LCD, then solve; CHECK every answer in the original — extraneous solutions are real.