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RF3 · Solve rational equations and reject extraneous solutions

Learn to solve rational equations and reject extraneous solutions through clear examples and targeted practice.

Alberta Grade 12 Mathematics

Relations and Functions

Find solutions, then check that they are allowed

A rational expression is a fraction whose numerator and denominator are polynomials. A rational equation contains one or more rational expressions. Solving one takes more than finding values through algebra. You must also check that each value is allowed in the original equation. A value that makes an original denominator zero is not a solution, even if it appears during the algebra.

What you will learn

  • Identify values that make an original denominator zero.
  • Solve rational equations by clearing denominators.
  • Check candidate values in the original equation.
  • Reject extraneous solutions and values that do not fit a stated context.

Review: fractions and restrictions

A fraction is undefined when its denominator is zero. Division by zero is not allowed. For example, in 5x−2\frac{5}{x-2}, the denominator is zero when x=2x=2. That value is called a non-permissible value for the expression.
Before solving a rational equation, set each original denominator equal to zero and find the excluded values. Keep the restrictions visible as you work. A value that makes an original denominator zero cannot be an answer.
You may already know how to solve an equation by doing the same operation to both sides. You may also know how to factor a quadratic and use the zero-product rule: if two factors multiply to zero, at least one factor must be zero. These familiar skills can help solve the equation after its fractions are cleared.
d≠0d\ne 0
  • An original denominator cannot equal zero.
  • Find restrictions before clearing denominators.

Clear denominators, then solve

The least common denominator, or LCD, is the smallest expression that each denominator divides into evenly. Multiply every term on both sides of the equation by the LCD. This removes the fractions, as long as you remember the original restrictions.
For example, a whole-number term still needs to be multiplied by the LCD. It can be thought of as a fraction with denominator 11. Apply the multiplication to every term, including terms on both sides.
Solve the resulting equation using familiar algebra. Each value you find is a candidate: a possible answer that still needs checking. Clearing denominators can make an equation easier to solve, but the new equation does not by itself show all the original restrictions.
each term×LCD\text{LCD}
  • Multiply every term by the LCD.
  • Keep the original restrictions while solving.
  • Treat values found after clearing denominators as candidates, not automatic answers.

Check and reject extraneous solutions

Substitute each candidate into the original equation. Reject it if it makes an original denominator zero, or if the original equation is not true. A candidate that fails the original equation is an extraneous solution. In a word problem, also check any stated conditions, such as a time needing to be positive.
A useful routine is: record restrictions, clear denominators, solve, and check every candidate in the original equation. This check matters because the original equation has restrictions that may be hidden once its denominators have been cleared.
  • Check candidates in the original equation, not only the cleared equation.
  • Reject values that make an original denominator zero.
  • Apply relevant conditions from the context.

A concrete check

For x+1x−2=3\frac{x+1}{x-2}=3, the denominator gives the restriction x≠2x\ne2. Multiplying both sides by x−2x-2 gives x+1=3(x−2)x+1=3(x-2). Solving gives x=72x=\frac{7}{2}. This candidate is allowed, and substituting it into the original equation gives a true statement.
Notice the purpose of the check: the restriction comes from the original fraction, while the candidate comes from solving. Keeping both in view helps you decide which values are genuine solutions.
x≠2x\ne2
  • Record the restriction before multiplying.
  • A valid answer must satisfy the original equation and its restrictions.

Worked example

A linear equation with a denominator restriction

Solve x+1x−2=3\frac{x+1}{x-2}=3.
  1. Record the restriction
    The denominator cannot be zero. Therefore, x=2x=2 is not allowed.
    x−2≠0⇒x≠2x-2\ne0\Rightarrow x\ne2
  2. Clear the denominator
    Multiply both sides by x−2x-2. This is valid for the allowed values, which exclude 22.
    x+1=3(x−2)x+1=3(x-2)
  3. Solve and check
    Solving gives x=72x=\frac{7}{2}. This value is not the restricted value 22. Substitution into the original equation gives a left side of 33, so the candidate works.
    x+1=3x−6⇒x=72x+1=3x-6\Rightarrow x=\frac{7}{2}
Answer: x=72x=\frac{7}{2}
Check: Substitution gives 72+172−2=3\frac{\frac{7}{2}+1}{\frac{7}{2}-2}=3.

Worked example

Two fractions and a quadratic

Solve 2x−1+1x+1=1\frac{2}{x-1}+\frac{1}{x+1}=1.
  1. Find restrictions and the LCD
    The original denominators are zero at x=1x=1 and x=−1x=-1. Exclude both values. The LCD is (x−1)(x+1)(x-1)(x+1).
    x≠1,x≠−1x\ne1,\quad x\ne-1
  2. Multiply every term by the LCD
    Multiplying every term by (x−1)(x+1)(x-1)(x+1) removes the denominators. The term 11 on the right must also be multiplied by the full LCD.
    2(x+1)+(x−1)=(x−1)(x+1)2(x+1)+(x-1)=(x-1)(x+1)
  3. Solve the resulting equation
    Expand and collect terms to get a quadratic equation. The quadratic formula gives two candidates.
    x2−3x−2=0⇒x=3±172x^2-3x-2=0\Rightarrow x=\frac{3\pm\sqrt{17}}{2}
  4. Check both candidates
    Neither candidate equals 11 or −1-1. Each satisfies the equation obtained after multiplying by the LCD. Since neither makes an original denominator zero, each also satisfies the original equation.
    x=3+172,x=3−172x=\frac{3+\sqrt{17}}{2},\quad x=\frac{3-\sqrt{17}}{2}
Answer: x=3+172x=\frac{3+\sqrt{17}}{2} or x=3−172x=\frac{3-\sqrt{17}}{2}
Check: Both candidates avoid the excluded values 11 and −1-1 and satisfy the original equation.

Worked example

A combined work-rate situation

Two pumps fill a tank together. One pump would take xx hours alone, and the other would take x+6x+6 hours alone. Working together, they fill the tank in 44 hours. Find xx.
  1. Write the equation and restrictions
    A pump that fills one tank in xx hours fills 1x\frac{1}{x} of the tank per hour. The combined hourly rate is the sum of the two rates. The times must be positive, so x>0x>0. The denominators also exclude x=0x=0 and x=−6x=-6.
    1x+1x+6=14\frac{1}{x}+\frac{1}{x+6}=\frac{1}{4}
  2. Clear the denominators
    Multiply every term by 4x(x+6)4x(x+6). This removes the fractions for values in the original equation's allowed domain.
    4(x+6)+4x=x(x+6)4(x+6)+4x=x(x+6)
  3. Solve and apply the context
    Rearranging gives a quadratic with candidates 66 and −4-4. The value −4-4 is not a possible time, so reject it. The value 66 is positive and does not make an original denominator zero. x^2-2x-24=0\Rightarrow x=6 or x=-4
  4. Check the accepted time
    At x=6x=6, the pumps take 66 hours and 1212 hours alone. Their rates add to one-quarter of a tank per hour, as required.
    16+112=14\frac{1}{6}+\frac{1}{12}=\frac{1}{4}
Answer: The first pump takes 66 hours to fill the tank alone.
Check: The second pump takes 1212 hours alone, and their combined rate is 14\frac{1}{4} tank per hour.

Common mistakes and how to avoid them

Multiplying by the LCD but forgetting a term that has no denominator written.
Correction: Multiply every term. An integer such as 11 is still a term and must be multiplied by the full LCD.
Keeping every value found after solving the cleared equation.
Correction: Check each candidate in the original equation. Reject values that make an original denominator zero or fail the equation.
Finding restrictions only after solving.
Correction: Find the non-permissible values from the original denominators before clearing them.
Accepting a negative time or another value that does not fit the situation.
Correction: Check the context as well as the algebra. A candidate may fail a stated condition even if it is not an excluded denominator value.

Lesson summary

  • Find values that make an original denominator zero, and exclude them.
  • Multiply every term by the LCD to clear the fractions.
  • Solve the resulting equation for candidate values.
  • Check candidates in the original equation and against relevant context conditions.
  • Report only values that pass every check.

Check your understanding

Question 1

For 3x+4=2\frac{3}{x+4}=2, which value is non-permissible?
  1. x=−4x=-4
  2. x=4x=4
  3. x=2x=2
  4. x=0x=0
Show answer and explanation
x=−4x=-4
The denominator x+4x+4 equals zero when x=−4x=-4, so that value is excluded.

Question 2

Solve 1x−2=1\frac{1}{x-2}=1. Which answer is valid?
  1. x=1x=1
  2. x=2x=2
  3. x=3x=3
  4. There is no solution
Show answer and explanation
x=3x=3
The restriction is x≠2x\ne2. Multiplying by x−2x-2 gives 1=x−21=x-2, so x=3x=3, which is allowed.

Question 3

A candidate solution makes an original denominator equal to zero. What should you do?
  1. Keep it if it solves the cleared equation.
  2. Reject it because the original expression is undefined.
  3. Round it to a nearby allowed value.
  4. Substitute it only into the cleared equation.
Show answer and explanation
Reject it because the original expression is undefined.
An original rational expression is undefined when its denominator is zero, so the candidate cannot be a solution.

Key terms

Rational expression
A fraction whose numerator and denominator are polynomials.
Non-permissible value
A value that makes a denominator zero, so the expression is undefined.
Least common denominator (LCD)
The smallest expression that each denominator divides into evenly.
Candidate
A value found while solving that still needs to be checked in the original equation.
Extraneous solution
A candidate produced during solving that does not satisfy the original equation or its restrictions.

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Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-2), expectation RF3. It is a study resource, not an official curriculum publication.

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