C3.6 Solve simple rational equations | Ontario Grade 12 Mathematics

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Use restrictions and a common denominator to find and check solutions

A rational equation is an equation that contains one or more fractions with a variable in a denominator. These equations can often be solved by clearing the denominators. The key is to note which values are not allowed before multiplying, and then check the answer in the original equation. This lesson focuses on simple rational equations that become linear equations after the denominators are cleared.

What you will learn

Recognize a rational equation and identify values that make its denominators zero.
Solve a simple rational equation by multiplying each term by the least common denominator.
Check a proposed solution in the original equation and reject any value that makes a denominator zero.

1. Review: fractions and excluded values

A denominator is the bottom part of a fraction. A fraction is undefined when its denominator is zero, so a denominator containing a variable places a restriction on that variable. For example, in

\frac{1}{x-4}

, the value

x=4

is not allowed because it makes the denominator zero.

To find a restriction, set each denominator equal to zero and solve. The values you find are excluded from the equation. Keep these restrictions in view throughout the solution. A value can solve the equation you get after clearing denominators and still be invalid if it was excluded in the original equation.

You also need the distributive property: multiplying a sum means multiplying every term in the sum. When clearing denominators, multiply every term on both sides, not just the fractions.

b=0

A variable value that makes an original denominator zero is not allowed.
Record restrictions before clearing denominators.
Apply multiplication to every term on both sides.

2. The method: clear denominators

The least common denominator, or LCD, is the smallest expression that each denominator divides into evenly. For example, the LCD of

x

and

x+2

is

x(x+2)

. Finding the LCD helps you clear all the fractions in one step.

Multiply every term in the equation by the LCD. Each denominator then cancels with its matching factor in the LCD. The resulting equation has no fractions, but it must still respect the restrictions from the original equation.

After clearing the denominators, solve the equation using familiar algebra steps. For a simple rational equation, the result is often a linear equation. Finally, substitute the proposed value into the original equation. This confirms that the value is allowed and makes both sides equal.

\text{LCD}

·each term

Find the LCD from all denominators in the equation.
Multiply every term by the LCD.
Solve the new equation, then check in the original equation.

3. Guided example and interpretation

Consider an equation with two fractions that have the same denominator. First identify the value that makes that denominator zero. Then multiply every term by the denominator. The fractions simplify, leaving an equation that can be solved with basic algebra.

The restriction is not an optional note. It tells us whether the final candidate can belong to the original equation. Substitution into the original equation provides a direct check: the denominators must be defined, and the two sides must have the same value.

\text{original equation}\;\longrightarrow\;\text{equation without fractions}

A shared denominator can be used as the LCD.
Cancelling a denominator is justified here by multiplying the entire equation by the LCD, which is nonzero for allowed values.
The check uses the original equation, not only the equation after clearing denominators.

4. A reliable final check

A solution is a value that makes the original equation true. Clearing denominators is a way to find candidates; it does not remove the need to check them. Substitute a candidate into each original denominator first. If any denominator becomes zero, reject that candidate.

If every denominator is defined, evaluate both sides. When the two sides are equal, the candidate is a solution. When they are not equal, review the algebra, especially the step where the LCD was distributed across terms.

This method is intended for simple rational equations that can be handled with algebra already familiar from earlier courses. Focus on the denominator restrictions, careful multiplication, and verification.

x∉\{\text{values that make a denominator }0\}

A candidate must be allowed in the original equation.
A valid solution makes the original left and right sides equal.
Do not check only the simplified equation.

The solving process at a glance

Stage	What to do	Purpose
Restrictions	Set each original denominator not equal to zero.	Identify values that cannot be solutions.
Clear fractions	Multiply every term by the LCD.	Create an equation without denominators.
Solve	Use algebra to find a candidate value.	Find a possible solution.
Check	Substitute into the original equation.	Confirm the value is allowed and both sides match.

Worked example

Clear a shared denominator

Solve

\frac{3}{x+1}=1+\frac{1}{x+1}

.

Find the restriction
The denominator is $x+1$ . It cannot equal zero, so $x$ cannot be $-1$ .
$x+1\ne 0 \Rightarrow x\ne -1$
Multiply by the LCD
The LCD is $x+1$ . Multiply every term on both sides by it. Since the restriction says $x+1$ is nonzero, this multiplication is valid.
$(x+1)(\frac{3}{x+1})=(x+1)(1)+(x+1)(\frac{1}{x+1})$
Simplify and solve
The matching factors cancel in each fraction. Solve the resulting linear equation by subtracting $2$ from both sides.
$3=x+1+1 \Rightarrow x=1$
Check in the original equation
The value $1$ is not excluded. Substitution gives a left side of $\frac{3}{2}$ and a right side of $1+\frac{1}{2}$ , which are equal.
$\frac{3}{1+1}=1+\frac{1}{1+1}$

Answer: The solution is

x=1

.

Check: The original denominator is

x+1

, which equals

2

at

x=1

, so the expression is defined. Both sides equal

\frac{3}{2}

.

Common mistakes and how to avoid them

Multiplying only the fraction terms by the LCD.

Correction: Multiply every term on both sides, including terms that are whole numbers.

Forgetting to record values that make a denominator zero.

Correction: Find restrictions from the original equation before multiplying, and compare every candidate with them.

Checking the answer only in the equation after clearing denominators.

Correction: Substitute into the original equation. The original denominators determine whether the value is allowed.

Lesson summary

Find and record the values that make an original denominator zero.
Find the LCD and multiply every term by it.
Solve the resulting equation using familiar algebra.
Check each candidate in the original equation.

Check your understanding

Question 1

What value must be excluded when solving

\frac{2}{x-5}=1

?

$x=2$
$x=5$
$x=-5$
correctIndex":1,"explanation":"The denominator is $x-5$ . Setting it equal to zero gives $x=5$ , so that value is excluded."

Show answer and explanation

x=5

The denominator is

x-5

. Setting it equal to zero gives

x=5

, so that value is excluded.

Question 2

What is the LCD for the denominators

x

and

x+3

?

$x+(x+3)$
$x(x+3)$
$3x$
correctIndex":1,"explanation":"The product $x(x+3)$ contains both denominator factors, so each denominator divides into it evenly."

Show answer and explanation

x(x+3)

The product

x(x+3)

contains both denominator factors, so each denominator divides into it evenly.

Question 3

Which statement about checking a candidate solution is correct?

Check it only in the equation after clearing denominators.
Check that it is allowed and makes both sides of the original equation equal.
Accept it whenever the algebra gives a number.
correctIndex":1,"explanation":"A candidate must not make an original denominator zero, and substitution into the original equation must give equal sides."

Show answer and explanation

Check that it is allowed and makes both sides of the original equation equal.

A candidate must not make an original denominator zero, and substitution into the original equation must give equal sides.

Key terms

Denominator: The bottom part of a fraction.
Rational equation: An equation containing one or more fractions with a variable in a denominator.
Restriction: A value that is not allowed because it makes an original denominator zero.
Least common denominator (LCD): The smallest expression that each denominator divides into evenly.
Solution: A value that makes the original equation true and is allowed in its denominators.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation C3.6. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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C3.6 · Solve simple rational equations

Ontario Grade 12 Mathematics

Use restrictions and a common denominator to find and check solutions

What you will learn

1. Review: fractions and excluded values

2. The method: clear denominators

3. Guided example and interpretation

4. A reliable final check

The solving process at a glance

Clear a shared denominator

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MHF4U

About this lesson and its review