Perform operations on two rational expressions

Alberta Grade 12 Mathematics

Relations and Functions

Adding, subtracting, multiplying, and dividing with attention to restrictions

A rational expression is a quotient of polynomials. For example,

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

is rational, and it is undefined when its denominator is zero. In this lesson, you will perform operations on two rational expressions. The central habits are to find restrictions first, use the correct operation rule, and check that simplification has not hidden a restriction.

What you will learn

Identify values that make a rational expression undefined.
Add or subtract rational expressions by using a common denominator.
Multiply or divide rational expressions and simplify the result.
State non-permissible values from the original expressions.

1. Review: fractions and restrictions

A denominator cannot equal zero because division by zero is undefined. To find a restriction, set each original denominator equal to zero and solve. Values found this way are called non-permissible values. They are excluded from the expression's domain, which is the set of allowed input values.

The same idea applies to rational expressions. For

\frac{x+4}{x+1}

, the denominator is zero when

x=-1

, so

x=-1

is not allowed. A factor may cancel later, but the original restriction remains. Cancellation changes the written form; it does not make an originally undefined value allowed.

For ordinary fractions, addition and subtraction need a common denominator. Multiplication uses numerators together and denominators together. Division by a fraction means multiplying by its reciprocal, which is the fraction with its numerator and denominator switched. These rules also apply to rational expressions.

b\ne 0

Find restrictions from the denominators before simplifying.
Keep original restrictions even if a factor cancels.

2. Add and subtract using a common denominator

To add or subtract rational expressions, first find a common denominator. A common denominator is an expression divisible by each denominator. Rewrite each fraction so it has that denominator, then add or subtract the numerators. Do not add or subtract the denominators.

If the denominators are already the same, keep that denominator and combine only the numerators. If they differ, factor them when possible. A common denominator can then be formed from the factors needed to include both denominators.

After combining, factor and simplify if possible. Simplifying a rational expression means dividing its numerator and denominator by the same non-zero factor. Record all restrictions from the original denominators.

\frac{A}{B}+\frac{C}{D}=\frac{AD+BC}{BD}

Use a common denominator before combining numerators.
Subtract the entire second numerator, using brackets when needed.
State restrictions from the original expressions.

3. Multiply and divide

To multiply, multiply the numerators and multiply the denominators. Factor first when possible. Then cancel factors that appear in both numerator and denominator. A factor can be cancelled only when it is a factor of the entire numerator and the entire denominator; terms joined by addition are not separate factors.

To divide by a rational expression, multiply by its reciprocal. The divisor must not equal zero. This adds a restriction: the numerator of the divisor cannot be zero. The divisor's own denominator also cannot be zero.

As with addition and subtraction, restrictions come from the original expressions. Check every original denominator, and, for division, also exclude values that make the divisor equal to zero. Give the simplified answer together with these restrictions.

\frac{A}{B}\div\frac{C}{D}=\frac{A}{B}·\frac{D}{C}

Factor before cancelling.
For division, multiply by the reciprocal of the second expression.
A divisor cannot equal zero.

4. A reliable method

A consistent order helps prevent errors. First write restrictions. Next identify the operation. For addition or subtraction, find a common denominator; for multiplication or division, use the matching rule. Then simplify and state the answer with its restrictions.

A final check should compare the answer with the original expressions. Make sure no original denominator could be zero, and confirm that division did not use a zero divisor. When subtracting, check the signs in the combined numerator.

Restrictions are part of the answer.
Simplify only after applying the correct operation.

Worked example

Add expressions with different denominators

Simplify

\frac{2}{x-1}+\frac{3}{x+2}

and state all non-permissible values.

Find restrictions
Each original denominator must be non-zero. The first excludes $x=1$ , and the second excludes $x=-2$ .
$x\ne 1, x\ne -2$
Use a common denominator
The denominators are different, so use their product, $(x-1)(x+2)$ . Multiply each numerator by the missing factor.
$\frac{2(x+2)+3(x-1)}{(x-1)(x+2)}$
Combine and simplify
Expand the numerator and combine like terms. The numerator does not share a factor with either denominator factor, so the expression is simplified.
$\frac{5x+1}{(x-1)(x+2)}$

Answer:

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

, where

x\ne 1,-2

.

Check: The common denominator contains both original denominator factors, so both original restrictions remain excluded.

Worked example

Multiply and cancel factors

Simplify

\frac{x^2-9}{x^2+4x+3}\cdot\frac{x+1}{x-3}

and state the restrictions.

Find restrictions
Factor the original denominators to see when they are zero. The first denominator has zeros at $x=-1$ and $x=-3$ ; the second has a zero at $x=3$ .
$x\ne -1, x\ne -3, x\ne 3$
Factor and multiply
Factor the difference of squares and the quadratic denominator. Multiplying the expressions gives the indicated factors in the numerator and denominator.
$\frac{(x-3)(x+3)(x+1)}{(x+1)(x+3)(x-3)}$
Cancel common factors
Each factor occurs in both the numerator and denominator, so they cancel. Keep the restrictions found from the original denominators.
$1$

Answer:

1

, where

x\ne -1,-3,3

.

Check: The simplified value is 1 only for inputs allowed in the original expression. The cancelled factors do not remove the original restrictions.

Worked example

Divide using a reciprocal

Simplify

\frac{x^2-4}{x+1}\div\frac{x-2}{x^2-1}

and state all restrictions.

Find restrictions
The first denominator excludes $x=-1$ . The second denominator excludes $x=-1$ and $x=1$ . Also, the divisor cannot be zero, so its numerator $x-2$ cannot equal zero.
$x\ne -1, x\ne 1, x\ne 2$
Multiply by the reciprocal
Switch the numerator and denominator of the second expression, then multiply. This is the rule for dividing by a fraction.
$\frac{x^2-4}{x+1}·\frac{x^2-1}{x-2}$
Factor and simplify
Factor the differences of squares. Cancel the common factors, but retain every restriction found from the original expressions.
$\frac{(x-2)(x+2)(x-1)(x+1)}{(x+1)(x-2)}=(x+2)(x-1)$

Answer:

(x+2)(x-1)

, where

x\ne -1,1,2

.

Check: The divisor is zero at

x=2

, so that value must be excluded even though the final simplified expression is defined there.

Common mistakes and how to avoid them

Adding the denominators when adding rational expressions.

Correction: Find a common denominator, then combine only the numerators.

Cancelling a term from a sum, such as cancelling one part of

x+3

against a denominator.

Correction: Factor first. Cancel only identical factors that multiply the entire numerator and denominator.

Forgetting a restriction after a factor cancels.

Correction: Keep the restrictions found from every original denominator.

Dividing without checking whether the divisor is zero.

Correction: Exclude values that make the divisor's numerator zero, as well as values that make any original denominator zero.

Lesson summary

Find non-permissible values from original denominators before simplifying.
For addition and subtraction, use a common denominator and combine the numerators.
For multiplication, multiply numerators and denominators, then cancel common factors.
For division, multiply by the reciprocal and exclude values that make the divisor zero.
State all restrictions in the final answer.

Check your understanding

Question 1

Simplify

\frac{1}{x}+\frac{2}{x+1}

. Which expression is correct?

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

Show answer and explanation

\frac{2x+3}{x(x+1)}

Using denominator

x(x+1)

gives numerator

(x+1)+2x=3x+1

. Correction: the matching option is

\frac{3x+1}{x(x+1)}

.

Question 2

Simplify

\frac{x^2-16}{x-4}

for allowed values of

x

.

$x+4$ , where $x\ne 4$
$x-4$ , where $x\ne 4$
$x+4$ , with no restriction
$x^2-16$ , where $x\ne 4$

Show answer and explanation

x+4

, where

x\ne 4

Factor the numerator as

(x-4)(x+4)

and cancel the common factor. The original denominator excludes

x=4

.

Question 3

What must be excluded when dividing

\frac{1}{x+2}

by

\frac{x-5}{x+1}

?

$x=-2$ only
$x=-1$ only
$x=-2,-1,5$
$x=2,1,5$

Show answer and explanation

x=-2,-1,5

The original denominators exclude

x=-2

and

x=-1

. The divisor is zero when its numerator

x-5

is zero, so exclude

x=5

as well.

Key terms

Rational expression: A quotient of polynomials, such as $\frac{x+1}{x-2}$ .
Non-permissible value: A value of the variable that makes an original denominator zero, or makes a divisor equal to zero.
Common denominator: A denominator that is a multiple of each denominator being combined.
Reciprocal: A fraction formed by switching the numerator and denominator.
Factor: A quantity multiplied by another quantity. In $(x-2)(x+3)$ , the factors are $x-2$ and $x+3$ .

Continue through Math 30-2

View the complete Math 30-2 Alberta Grade 12 Mathematics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-2), expectation RF2. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question

RF2 · Perform operations on two rational expressions

Alberta Grade 12 Mathematics

Adding, subtracting, multiplying, and dividing with attention to restrictions

What you will learn

1. Review: fractions and restrictions

2. Add and subtract using a common denominator

3. Multiply and divide

4. A reliable method

Add expressions with different denominators

Multiply and cancel factors

Divide using a reciprocal

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through Math 30-2

About this lesson and its review