1. What is a rational expression
An expression of the form $\dfrac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomials and $Q$ is not the zero polynomial.
The name is exactly right: a rational number is a ratio of integers; a rational expression is a ratio of polynomials. Numerator and denominator can each be as simple as a constant or as elaborate as $x^4 - 5x^2 + 6$. Here are a few:
$$ \frac{1}{x}, \qquad \frac{x+3}{x-2}, \qquad \frac{x^2 - 4}{x^2 - x - 6}, \qquad \frac{5}{x^2 + 1}. $$Every move you make on rational expressions — simplifying, multiplying, adding, solving — is a move on the polynomials inside. So everything you know about factoring pays off immediately and constantly here.
Think of a rational expression as a numerical fraction in disguise: each $x$ is just a number you haven't picked yet. Anything that's true of $\tfrac{6}{8}$ — that it simplifies to $\tfrac{3}{4}$, that adding $\tfrac{1}{3}$ to it needs a common denominator — is true of $\tfrac{P(x)}{Q(x)}$. The only new wrinkle is that some values of $x$ are forbidden.
2. Domain restrictions
You learned long ago that you can't divide by zero. Nothing about that rule weakens when the divisor is an expression. The values of $x$ that make the denominator equal to zero are simply not allowed:
For $\dfrac{P(x)}{Q(x)}$, the restricted values are all $x$ satisfying $Q(x) = 0$. The domain is every real number except the restricted values.
To find them, you set the denominator equal to zero and solve. That's it.
| Expression | Denominator | Restricted values |
|---|---|---|
| $\dfrac{1}{x}$ | $x$ | $x = 0$ |
| $\dfrac{x+3}{x-2}$ | $x-2$ | $x = 2$ |
| $\dfrac{x+1}{x^2 - 4}$ | $(x-2)(x+2)$ | $x = 2,\ x = -2$ |
| $\dfrac{5}{x^2 + 1}$ | $x^2 + 1$ | none (never zero for real $x$) |
If you simplify an expression and the simplification removes a factor from the denominator, the original restriction still applies. $\dfrac{x^2 - 4}{x - 2}$ "simplifies" to $x + 2$ — but the original is undefined at $x = 2$, while the simplified form is perfectly happy there. The two are equal for $x \neq 2$, not for all $x$.
3. Simplifying by factoring
To simplify $\tfrac{6}{8}$, you factor: $\tfrac{6}{8} = \tfrac{2 \cdot 3}{2 \cdot 4} = \tfrac{3}{4}$. The same idea works here. The recipe is two steps:
- Factor numerator and denominator completely.
- Cancel any factor that appears top and bottom.
For example,
$$ \frac{x^2 - 9}{x^2 - x - 6} \;=\; \frac{(x-3)(x+3)}{(x-3)(x+2)} \;=\; \frac{x+3}{x+2} \quad (x \neq 3,\ x \neq -2). $$Notice the annotation. Both $x = 3$ and $x = -2$ make the original denominator zero, so both stay out of the domain even though the $(x-3)$ has disappeared from the page.
The cancellation rule, stated carefully
You can cancel a common factor from numerator and denominator — that is, something multiplied by everything else on its side. You cannot cancel a common term that is added to or subtracted from something else.
That distinction is the single most common source of errors in this whole topic, so look at it carefully:
- $\dfrac{3x}{3y} = \dfrac{x}{y}$ — $3$ is a factor of both
- $\dfrac{(x+1)(x-2)}{(x+1)(x+3)} = \dfrac{x-2}{x+3}$
- $\dfrac{x^2 - 4}{x - 2} = \dfrac{(x-2)(x+2)}{x-2} = x+2$
- $\dfrac{x + 3}{x} \neq 3$ — the $x$ is a term, not a factor of the whole numerator
- $\dfrac{x + 3}{x + 5} \neq \dfrac{3}{5}$ — the $x$'s aren't factors
- $\dfrac{x^2 + 1}{x} \neq x + 1$ — only $x^2$ has an $x$ factor; the $+1$ doesn't
If you're not sure whether something is a factor or a term, look at the operation between it and its neighbors. Multiplication and division separate factors; addition and subtraction separate terms. Only factors cancel.
When tempted to cancel something that feels suspicious, pick a small value of $x$ (say $x = 1$) and evaluate both sides. If they disagree, the cancellation was illegal. $\dfrac{1 + 3}{1} = 4$, but $3$ is $3$. End of debate.
4. Multiplying and dividing
Multiplication is the easiest operation. The rule is the same as for numerical fractions:
$$ \frac{A}{B} \cdot \frac{C}{D} \;=\; \frac{A \cdot C}{B \cdot D}. $$In practice you almost never multiply out — that just creates an unfactored mess that you have to factor again to simplify. Better: factor everything first, write one big fraction, and cancel before multiplying.
$$ \frac{x+2}{x-3} \cdot \frac{x-3}{x^2 - 4} \;=\; \frac{(x+2)(x-3)}{(x-3)(x-2)(x+2)} \;=\; \frac{1}{x-2} \quad (x \neq -2,\,2,\,3). $$Division is just multiplication by the reciprocal:
$$ \frac{A}{B} \div \frac{C}{D} \;=\; \frac{A}{B} \cdot \frac{D}{C} \;=\; \frac{AD}{BC}. $$Flip the second fraction, then proceed as in multiplication. For example,
$$ \frac{x^2}{x-1} \;\div\; \frac{x}{x-1} \;=\; \frac{x^2}{x-1} \cdot \frac{x-1}{x} \;=\; \frac{x^2(x-1)}{x(x-1)} \;=\; x \quad (x \neq 0,\,1). $$When you multiply or divide, the restricted values of the result include the restrictions of every factor that appeared along the way, not just the ones that survive simplification. In the multiplication above, $(x-3)$ cancelled — but $x = 3$ is still excluded, because the original expression was undefined there.
5. Adding and subtracting
You can't add fractions with different denominators directly. Numerical fractions are the same: $\tfrac{1}{2} + \tfrac{1}{3}$ isn't $\tfrac{2}{5}$ — you need a common denominator first.
The procedure has four steps:
- Factor every denominator.
- Find the least common denominator (LCD): for each distinct factor, take the highest power that appears in any denominator.
- Rewrite each fraction with the LCD by multiplying its numerator and denominator by whatever is missing.
- Add (or subtract) the numerators; keep the common denominator. Simplify.
Same denominator — easy case
$$ \frac{x}{x-1} \;-\; \frac{1}{x-1} \;=\; \frac{x - 1}{x - 1} \;=\; 1 \quad (x \neq 1). $$Different denominators
Take $\dfrac{1}{x} + \dfrac{1}{x+1}$. Both denominators are already prime; the LCD is just their product, $x(x+1)$.
$$ \frac{1}{x} + \frac{1}{x+1} \;=\; \frac{x+1}{x(x+1)} + \frac{x}{x(x+1)} \;=\; \frac{2x+1}{x(x+1)}. $$Shared factors — where the LCD pays off
When the denominators share factors, the LCD is smaller than the product. Consider
$$ \frac{3}{x^2 - 1} \;+\; \frac{2}{x + 1}. $$Factoring: $x^2 - 1 = (x-1)(x+1)$. The LCD is $(x-1)(x+1)$ — we don't need a second copy of $(x+1)$. So
$$ \frac{3}{(x-1)(x+1)} + \frac{2(x-1)}{(x-1)(x+1)} \;=\; \frac{3 + 2(x-1)}{(x-1)(x+1)} \;=\; \frac{2x + 1}{(x-1)(x+1)}. $$$\dfrac{a}{b} + \dfrac{c}{d} \neq \dfrac{a+c}{b+d}$. Adding numerators and denominators is one of the most stubborn beginner mistakes in algebra — and it gives wrong answers every single time. If the denominators differ, you must get a common one first.
6. Complex fractions
A complex fraction is a fraction whose numerator or denominator (or both) contains a fraction. Like
$$ \frac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}}. $$You have two ways to simplify it. Both work; pick whichever you find easier.
Method 1 — multiply by the LCD of the inner fractions
Find the LCD of every embedded fraction — in this case, just $x$ — and multiply the main numerator and denominator by it. The internal fractions vanish in one step:
$$ \frac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}} \;\cdot\; \frac{x}{x} \;=\; \frac{x + 1}{x - 1} \quad (x \neq 0,\,1). $$Method 2 — combine, then divide
Turn the main numerator and main denominator into single fractions, then divide (flip and multiply):
$$ \frac{\frac{x+1}{x}}{\frac{x-1}{x}} \;=\; \frac{x+1}{x} \cdot \frac{x}{x-1} \;=\; \frac{x+1}{x-1}. $$Same answer. Method 1 is usually faster when there are many internal fractions; Method 2 is cleaner when the numerator and denominator each combine in one step.
If you use Method 1, the LCD multiplies both the main numerator and the main denominator — you're really multiplying the whole fraction by $\tfrac{x}{x} = 1$. Multiplying only the top or only the bottom changes the value.
7. Solving rational equations
An equation containing rational expressions is called a rational equation. Solving one almost always starts the same way: clear the denominators by multiplying both sides by the LCD. That converts the rational equation into a polynomial equation, which you already know how to handle.
But this move comes with a hidden cost. Multiplying both sides by an expression that can be zero for some value of $x$ can introduce false solutions — values that satisfy the cleared equation but make the original equation undefined. These are called extraneous solutions, and you must check for them every time.
A solution of the cleared-denominator equation that is not in the domain of the original equation. It must be discarded.
The procedure
- Find the restricted values of every denominator in the original equation.
- Multiply both sides by the LCD to clear all denominators.
- Solve the resulting polynomial equation.
- Discard any candidate equal to a restricted value — those are extraneous.
- Verify the survivors in the original equation.
A clean case
Solve $\dfrac{1}{x} + \dfrac{1}{x+1} = \dfrac{1}{2}$.
Restricted values: $x \neq 0$ and $x \neq -1$. LCD: $2x(x+1)$. Multiply both sides:
$$ 2(x+1) + 2x \;=\; x(x+1). $$Expand and rearrange:
$$ 4x + 2 \;=\; x^2 + x \quad\Longrightarrow\quad x^2 - 3x - 2 \;=\; 0. $$By the quadratic formula, $x = \tfrac{3 \pm \sqrt{17}}{2}$. Neither value is $0$ or $-1$, so both are valid.
A case where checking matters
Solve $\dfrac{1}{x-2} = \dfrac{2}{x^2 - 4}$.
Restricted values: $x^2 - 4 = (x-2)(x+2) = 0$ gives $x = 2$ and $x = -2$. Both are excluded. LCD: $(x-2)(x+2)$. Multiply both sides:
$$ (x+2) \;=\; 2 \quad\Longrightarrow\quad x \;=\; 0. $$Check: $x = 0$ is not a restricted value, and substituting back into the original gives $\tfrac{1}{-2} = \tfrac{2}{-4}$, both equal to $-\tfrac{1}{2}$. Valid.
Had the algebra produced $x = 2$ or $x = -2$, those would be extraneous — solutions of the cleared equation but undefined in the original.
Clearing denominators is the one move in rational algebra that does not preserve equivalence — it can manufacture solutions out of thin air. The check is not bookkeeping; it's the only thing standing between you and a wrong answer.
8. Common pitfalls
$\dfrac{x + 3}{x}$ does not simplify to $3$. The $x$ in the numerator is added to $3$, not multiplied — it isn't a factor of the whole top, so it doesn't cancel. The honest simplification is $1 + \dfrac{3}{x}$.
$\dfrac{x^2 - 4}{x-2}$ simplifies to $x+2$ — but the original was undefined at $x = 2$, and that restriction persists. The two expressions are equal only on the domain of the original. If a later step (a limit, a graph, an equation) asks about $x = 2$, the restriction matters.
$\dfrac{a}{b} + \dfrac{c}{d} \neq \dfrac{a+c}{b+d}$. You need a common denominator first. This rule has no exceptions; it does not become true for any choice of letters or expressions.
Multiplying both sides of a rational equation by an expression containing $x$ can introduce false solutions. Always verify every candidate against the restricted values of the original equation — and substitute back to confirm.
The sign of an expression like $x - 3$ depends on $x$, so multiplying an inequality by it may or may not flip the direction. Don't do it. For rational inequalities, move everything to one side and use a sign chart instead.
9. Worked examples
Try each one yourself before opening the solution. The goal isn't to memorize the answer — it's to confirm your moves line up with the procedure.
Example 1 · Find the restricted values of $\dfrac{x+1}{x^2 - x - 6}$
Step 1. Factor the denominator:
$$ x^2 - x - 6 \;=\; (x - 3)(x + 2). $$Step 2. Set each factor equal to zero:
$$ x - 3 = 0 \;\Rightarrow\; x = 3, \qquad x + 2 = 0 \;\Rightarrow\; x = -2. $$The restricted values are $x = 3$ and $x = -2$.
Example 2 · Simplify $\dfrac{x^2 - 4}{x^2 + 4x + 4}$
Step 1. Factor both:
$$ \frac{x^2 - 4}{x^2 + 4x + 4} \;=\; \frac{(x-2)(x+2)}{(x+2)(x+2)}. $$Step 2. Cancel the common factor $(x+2)$:
$$ = \;\frac{x-2}{x+2} \quad (x \neq -2). $$Note: $x = -2$ stays restricted even though $(x+2)$ no longer appears.
Example 3 · Add $\dfrac{2}{x-1} + \dfrac{3}{x+2}$
Step 1. The denominators share no factors; LCD $= (x-1)(x+2)$.
Step 2. Rewrite each fraction:
$$ \frac{2(x+2)}{(x-1)(x+2)} + \frac{3(x-1)}{(x-1)(x+2)}. $$Step 3. Combine numerators and simplify:
$$ \frac{2x + 4 + 3x - 3}{(x-1)(x+2)} \;=\; \frac{5x + 1}{(x-1)(x+2)} \quad (x \neq 1, -2). $$Example 4 · Simplify the complex fraction $\dfrac{\frac{1}{x} - \frac{1}{y}}{\frac{1}{x} + \frac{1}{y}}$
Step 1. The LCD of the inner fractions is $xy$. Multiply top and bottom by $xy$:
$$ \frac{\left(\frac{1}{x} - \frac{1}{y}\right) \cdot xy}{\left(\frac{1}{x} + \frac{1}{y}\right) \cdot xy} \;=\; \frac{y - x}{y + x}. $$Valid for $x \neq 0$, $y \neq 0$, and $x \neq -y$ (so the denominator doesn't vanish).
Example 5 · Solve $\dfrac{3}{x-1} - \dfrac{1}{x+1} = \dfrac{2}{x^2 - 1}$
Step 1. Factor: $x^2 - 1 = (x-1)(x+1)$. Restricted values: $x \neq 1$ and $x \neq -1$.
Step 2. LCD $= (x-1)(x+1)$. Multiply both sides:
$$ 3(x+1) - (x-1) \;=\; 2. $$Step 3. Simplify:
$$ 3x + 3 - x + 1 \;=\; 2 \quad\Longrightarrow\quad 2x + 4 = 2 \quad\Longrightarrow\quad x = -1. $$Step 4. Check: $x = -1$ is a restricted value. Extraneous. There is no solution.
Example 6 · Work-rate. Alice paints a room in 3 h, Bob in 4 h. How long working together?
Alice's rate is $\tfrac{1}{3}$ rooms/hour; Bob's is $\tfrac{1}{4}$. Their combined rate is
$$ \frac{1}{3} + \frac{1}{4} \;=\; \frac{4 + 3}{12} \;=\; \frac{7}{12} \;\text{rooms/hour}. $$Time to paint one room together:
$$ t \;=\; \frac{1}{7/12} \;=\; \frac{12}{7} \;\text{hours} \;\approx\; 1\,\text{h}\,43\,\text{min}. $$The combined-rate formula $\tfrac{1}{t} = \sum \tfrac{1}{t_i}$ is just rational-expression addition in disguise.