---
title: Rational Expressions and Equations
subtitle: Structure, restrictions, operations, graphs, and applications
author: Mathematics Department
organization: Your School
recipient: Algebra II students
date: 2026-09-13
version: "1.0"
description: A proof-led Algebra II lesson on rational expressions, domain restrictions, operations, equations, functions, and applications.
keywords: [rational expressions, rational equations, domain restrictions, extraneous solutions, asymptotes, Algebra II]
template: assignment
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header: Rational Expressions | Structure Before Procedure
footer: Student lesson and practice
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language: en-US
---

# Rational Expressions and Equations

A **rational expression** is a quotient

$$
R(x)=\frac{p(x)}{q(x)},
$$

where $p$ and $q$ are polynomials and $q$ is not the zero polynomial. Its domain excludes every value for which $q(x)=0$. The restriction is part of the expression's meaning: algebra may change its appearance, but it must not silently restore an input at which the original expression was undefined.

For example,

$$
\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3,
\qquad x\ne3.
$$

The simplified rule gives the same output as the original at every allowed input. It does not give the original expression a value at $x=3$.

## Learning goals

By the end of this lesson, you should be able to:

- determine and preserve domain restrictions;
- simplify by factoring and canceling factors, not terms;
- multiply, divide, add, and subtract rational expressions;
- simplify complex rational expressions;
- use polynomial division to expose end behavior;
- solve rational equations and reject excluded candidates;
- identify holes, intercepts, and vertical, horizontal, or slant asymptotes;
- build and interpret rational models involving rates and inverse variation.

## 1. Equality requires a domain

Two expressions are equivalent on a stated domain when they have equal values at every input in that domain. If their natural domains differ, writing them as unqualified equal functions is false.

### The cancellation theorem

Suppose $a(x)$, $b(x)$, and $c(x)$ are polynomials. For every $x$ such that $b(x)c(x)\ne0$,

$$
\frac{a(x)c(x)}{b(x)c(x)}=\frac{a(x)}{b(x)}.
$$

The proof is the ordinary multiplicative-inverse law:

$$
\frac{a(x)c(x)}{b(x)c(x)}
=\frac{a(x)}{b(x)}\cdot\frac{c(x)}{c(x)}
=\frac{a(x)}{b(x)}\cdot1.
$$

The condition $c(x)\ne0$ is essential because $c(x)/c(x)$ is then, and only then, equal to $1$. Cancellation removes a common **factor** from the notation; it does not erase the original excluded value.

:::figure{#fig:hole caption="Canceling a common factor preserves the missing input as a hole" alt="Coordinate graph of y equals x plus 1 with an open circle at the excluded point 1 comma 2, representing the rational function x squared minus 1 over x minus 1" width=86 placement=none}
![The line y equals x plus 1 with a hole at 1 comma 2](removable-discontinuity.svg)
:::

### Worked example 1: factor before reducing

Simplify and state the domain:

$$
\frac{x^2+x-6}{x^2-4}.
$$

Factoring gives

$$
\frac{(x+3)(x-2)}{(x-2)(x+2)}
=\frac{x+3}{x+2},
\qquad x\ne2,-2.
$$

The factor $x-2$ cancels, but $x=2$ remains excluded. The graph has a hole there. The uncanceled denominator factor $x+2$ produces a vertical asymptote at $x=-2$.

### Terms do not cancel across addition

The expression $(x+4)/x$ is not $4$. Division distributes over addition as

$$
\frac{x+4}{x}=\frac{x}{x}+\frac4x=1+\frac4x,
\qquad x\ne0.
$$

There is no common factor $x$ in the entire numerator. Factoring is the test for legal cancellation.

## 2. Multiplication and division

Rational expressions behave like rational numbers on inputs where all required denominators and divisors are nonzero.

$$
\frac ab\cdot\frac cd=\frac{ac}{bd},
\qquad
\frac ab\div\frac cd=\frac ab\cdot\frac dc.
$$

For division, $c/d$ itself must be nonzero, so $c\ne0$ in addition to $b,d\ne0$.

### Worked example 2: multiply after factoring

$$
\frac{x^2-9}{x^2-4}\cdot\frac{x+2}{x-3}
=\frac{(x-3)(x+3)}{(x-2)(x+2)}\cdot\frac{x+2}{x-3}
=\frac{x+3}{x-2}.
$$

The original expression requires

$$
x\ne-2,2,3.
$$

All three restrictions survive simplification.

### Worked example 3: division adds a restriction

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

The result is $1$ only for $x\ne-1,1$. At $x=-1$, denominators vanish. At $x=1$, the divisor equals zero, and division by zero is undefined.

## 3. Addition and subtraction

Unlike multiplication, addition requires a common denominator:

$$
\frac ab+\frac cd=\frac{ad+bc}{bd}.
$$

The least common denominator, built from each distinct factor at its greatest required power, usually minimizes later work.

### Worked example 4: unlike linear denominators

$$
\frac3x+\frac2{x+1}
=\frac{3(x+1)+2x}{x(x+1)}
=\frac{5x+3}{x(x+1)},
\qquad x\ne0,-1.
$$

### Worked example 5: repeated and distinct factors

Simplify

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

The denominators are $x(x-1)$ and $x(x+1)$, so the LCD is $x(x-1)(x+1)$. Therefore

$$
\frac{2(x+1)-(x-1)}{x(x-1)(x+1)}
=\frac{x+3}{x(x-1)(x+1)},
\qquad x\ne-1,0,1.
$$

## 4. Complex rational expressions

A **complex rational expression** contains a rational expression in its numerator, denominator, or both. It can be simplified by combining the numerator and denominator separately or by multiplying every term by their common LCD.

### Worked example 6: two variables

$$
\frac{\frac1x+\frac1y}{\frac1x-\frac1y}
=\frac{xy\left(\frac1x+\frac1y\right)}{xy\left(\frac1x-\frac1y\right)}
=\frac{x+y}{y-x}.
$$

The original requires $x\ne0$, $y\ne0$, and $x\ne y$ because its entire denominator must not equal zero.

### Worked example 7: a hidden excluded value

$$
\frac{1-\frac4{x^2}}{1+\frac2x}
=\frac{(x-2)(x+2)/x^2}{(x+2)/x}
=\frac{x-2}{x},
\qquad x\ne0,-2.
$$

Although $x+2$ disappears, $x=-2$ made the original complex denominator zero and remains excluded.

## 5. Polynomial division

If $p(x)$ and nonzero $q(x)$ are polynomials, there are unique polynomials $s(x)$ and $r(x)$ such that

$$
p(x)=q(x)s(x)+r(x),
\qquad \deg r<\deg q.
$$

Dividing by $q(x)$ gives

$$
\frac{p(x)}{q(x)}=s(x)+\frac{r(x)}{q(x)}.
$$

This is the polynomial counterpart of $13/3=4+1/3$. It also reveals the graph's end behavior because $r(x)/q(x)$ often approaches zero as $|x|$ grows.

### Worked example 8: quotient plus remainder

Divide $2x^3+3x^2-x+5$ by $x+2$. Polynomial division gives

$$
2x^3+3x^2-x+5=(x+2)(2x^2-x+1)+3.
$$

Hence

$$
\frac{2x^3+3x^2-x+5}{x+2}
=2x^2-x+1+\frac3{x+2},
\qquad x\ne-2.
$$

## 6. Rational equations

A rational equation contains at least one rational expression. Multiplying every term by the LCD is valid only on the original domain, where that LCD is nonzero. The resulting polynomial equation may contain an excluded candidate, so every candidate must be checked against the original restrictions.

### Worked example 9: two valid solutions

Solve

$$
\frac1x+\frac1{x+2}=\frac34,
\qquad x\ne0,-2.
$$

Multiplying by $4x(x+2)$ gives

$$
4(x+2)+4x=3x(x+2),
$$

so

$$
3x^2-2x-8=0
\quad\Longrightarrow\quad
(3x+4)(x-2)=0.
$$

Both $x=-4/3$ and $x=2$ are allowed and satisfy the original equation.

### Worked example 10: an excluded candidate

Solve

$$
\frac{x}{x-3}=\frac3{x-3}+2,
\qquad x\ne3.
$$

Multiplication by $x-3$ gives

$$
x=3+2(x-3),
$$

whose only candidate is $x=3$. That value is outside the original domain, so the equation has **no solution**. Clearing denominators produced a necessary algebraic candidate, not permission to ignore the denominator.

### Worked example 11: a radical answer is still exact

$$
\frac1{x-1}+\frac1{x+1}=1,
\qquad x\ne-1,1.
$$

After multiplying by $(x-1)(x+1)$,

$$
(x+1)+(x-1)=x^2-1,
$$

so $x^2-2x-1=0$ and

$$
x=1\pm\sqrt2.
$$

Neither value is excluded.

## 7. Rational functions and graph structure

Factoring and division expose different graph features.

- A canceled denominator zero is a **removable discontinuity**, usually shown as a hole.
- An uncanceled denominator zero is a candidate **vertical asymptote**.
- Equal numerator and denominator degrees give a horizontal asymptote equal to the ratio of leading coefficients.
- If the numerator degree is one more, polynomial division may reveal a slant asymptote.
- Zeros of the reduced numerator are $x$-intercepts; $f(0)$ is the $y$-intercept when defined.

:::figure{#fig:features caption="Factoring locates intercepts and vertical asymptotes; leading terms determine end behavior" alt="Coordinate graph of y equals x plus 3 over x minus 2 with vertical asymptote x equals 2, horizontal asymptote y equals 1, x intercept negative 3, and y intercept negative three halves" width=90 placement=none}
![Graph features of y equals x plus 3 over x minus 2](rational-function-features.svg)
:::

### Worked example 12: read every principal feature

For

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

the domain excludes $2$, the vertical asymptote is $x=2$, and the horizontal asymptote is $y=1$. The intercepts are $(-3,0)$ and $(0,-3/2)$. Division gives

$$
f(x)=1+\frac5{x-2},
$$

which makes both asymptotes and the translated reciprocal shape visible.

### Worked example 13: slant asymptote

$$
\frac{x^2+1}{x-1}=x+1+\frac2{x-1}.
$$

As $|x|$ grows, $2/(x-1)$ approaches $0$, so the graph approaches the line $y=x+1$. The vertical asymptote is $x=1$.

## 8. Rational models

Restrictions in a model include both algebraic exclusions and contextual constraints such as positive time, speed, or quantity.

### Worked example 14: combined work

One pump fills a tank in $6$ hours and another in $4$ hours. If $t$ is the time when both work together, their completed fractions add:

$$
\frac{t}{6}+\frac{t}{4}=1.
$$

Equivalently, $1/6+1/4=1/t$, so

$$
t=\frac{12}{5}=2.4\text{ hours}.
$$

The answer is less than either individual time, as the context requires.

### Worked example 15: round-trip speed

A driver travels $120$ miles out at speed $v$ and returns at $v+20$ miles per hour. A total time of $5$ hours gives

$$
\frac{120}{v}+\frac{120}{v+20}=5,
\qquad v>0.
$$

Clearing denominators yields $v^2-28v-480=0$, so $(v-40)(v+12)=0$. The algebraic candidates are $40$ and $-12$, but only $v=40$ mph is contextually meaningful.

## 9. Error analysis

Before accepting a result, ask:

1. Were all original denominator zeros listed before simplification?
2. Were only common factors canceled?
3. Did addition use a common denominator?
4. Did division exclude inputs that make the divisor zero?
5. Was every equation candidate checked against the original domain?
6. Does a model satisfy units, positivity, and feasibility?

> [!WARNING]
> An expression can be algebraically reduced while its domain is incorrectly enlarged. A solution can satisfy the cleared equation while remaining undefined in the original. These are logical errors, not merely notation errors.

# Practice set

Show restrictions before simplifying. Give exact values unless an approximation is requested.

## A. Domain, structure, and equivalence

1. State the domain of $\dfrac{x^2-9}{x^2+x-6}$.
2. Simplify $\dfrac{x^2-3x}{x^2-9}$ and preserve all original restrictions.
3. Explain why $\dfrac{x^2-9}{x-3}$ and $x+3$ are not the same function on their natural domains.
4. Simplify $\dfrac{x^2-4}{x-2}$ and identify the missing point on its graph.
5. Evaluate $\dfrac{2x}{x^2-1}$ at $x=2$.
6. Classify $\dfrac{3x^3-x}{x^2+1}$ as proper or improper.
7. Write $\dfrac{2x^3+3x^2-x+5}{x+2}$ as a polynomial plus a proper rational expression.
8. Find $k$ so that $\dfrac{x^2+kx-6}{x-2}$ has a removable discontinuity at $x=2$.

## B. Operations and complex fractions

9. Simplify $\dfrac{x^2-9}{x^2-4}\cdot\dfrac{x+2}{x-3}$.
10. Simplify $\dfrac{x^2-1}{(x+1)^2}\div\dfrac{x-1}{x+1}$.
11. Add $\dfrac3x+\dfrac2{x+1}$.
12. Subtract $\dfrac2{x-3}-\dfrac1{x+3}$.
13. Add $\dfrac{x}{x-2}+\dfrac4{x^2-4}$.
14. Simplify $\dfrac{\frac1x+\frac1y}{\frac1x-\frac1y}$.
15. Simplify $\dfrac{1-\frac4{x^2}}{1+\frac2x}$.
16. Subtract $\dfrac1{x^2-x}-\dfrac1{x^2+x}$.

## C. Rational equations

17. Solve $\dfrac1x=\dfrac16$.
18. Solve $\dfrac2{x-1}=\dfrac3{x+2}$.
19. Solve $\dfrac1x+\dfrac1{x+2}=\dfrac34$.
20. Solve $\dfrac{x+1}{x-2}=4$.
21. Solve $\dfrac{x}{x-3}=\dfrac3{x-3}+2$.
22. Solve $\dfrac{x^2-4}{x-2}=6$.
23. Solve $\dfrac1{x-1}+\dfrac1{x+1}=1$.
24. Find all real solutions of $\dfrac3{x+1}-\dfrac2{x-1}=1$.
25. Solve $\dfrac2x+\dfrac3{x-2}=1$.
26. Solve $\dfrac{x-2}{x-2}=0$.

## D. Rational-function structure

27. For $f(x)=\dfrac{x+3}{x-2}$, give the domain, intercepts, and vertical and horizontal asymptotes.
28. For $g(x)=\dfrac{x^2-9}{x-3}$, identify the simplified rule, domain, and hole.
29. For $h(x)=\dfrac{2x^2+1}{x^2-4}$, give the real intercepts and vertical and horizontal asymptotes.
30. For $p(x)=\dfrac1{x+1}-2$, give the domain, intercepts, and asymptotes.
31. Use division to find the slant asymptote of $q(x)=\dfrac{x^2+1}{x-1}$.
32. Construct a rational function with a vertical asymptote at $x=4$, a hole at $x=-1$, an $x$-intercept at $x=2$, and horizontal asymptote $y=1$.

## E. Models and reasoning

33. Pumps filling one tank alone in $6$ hours and $4$ hours work together. Find the filling time.
34. A $120$-mile outbound trip is driven at $v$ mph and the return at $v+20$ mph. The total time is $5$ hours. Find both speeds.
35. The variables satisfy inverse variation. If $y=6$ when $x=4$, find $x$ when $y=3$.
36. A thin-lens model is $\dfrac1f=\dfrac1{d_o}+\dfrac1{d_i}$. Find $d_i$ when $f=10$ cm and $d_o=30$ cm.
37. A print run has fixed setup cost $\$240$ and variable cost $\$6$ per copy. Find the average cost at $80$ copies and the number of copies required for an average cost of $\$7.50$.
38. Two resistors of $12\ \Omega$ and $18\ \Omega$ are connected in parallel. Use $\dfrac1R=\dfrac1{R_1}+\dfrac1{R_2}$ to find $R$.
39. A student writes $\dfrac{x+4}{x}=4$. Identify the invalid move and give a correct equivalent form.
40. Explain why canceling $x-2$ in $\dfrac{(x-2)(x+5)}{(x-2)(x-1)}$ does not permit $x=2$, and distinguish the graph feature at $x=2$ from the feature at $x=1$.

## References

- Common Core State Standards for Mathematics, HSA-APR.D.6-7 and HSA-REI.A.2.
- OpenStax, *Intermediate Algebra 2e*, Chapter 7, CC BY 4.0.
- Ankit Kumar Chauhan, *Dividing Polynomials*, Matherama, author-owned source consulted for polynomial division and the quotient-remainder form.

This packet uses real-valued Algebra II conventions. A teacher may extend selected equations to complex solutions, but multivariable rational functions, partial fractions, and calculus-based limit proofs lie outside its scope.