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Algebra 2 Rational functions

Solving rational equations (incl. extraneous solutions)

20 practice questions 0 video lessons Theory + worked examples

Solving Rational Equations

Common Core Algebra 2 • Standard A-REI.2 • Rational Functions

Solving Rational Equations is a topic in Rational Functions in the Common Core State Standards. It is aligned to Standard A-REI.2, which requires students to solve rational equations in one variable and give rise to and identify extraneous solutions.

Rational equations are solved by clearing denominators with the LCD, then rejecting extraneous solutions that make a denominator zero.

Common Core Algebra 2 › Rational Functions › Solving Rational Equations  —  Standard A-REI.2

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Theory

A rational equation has variables in a denominator. To solve:

  1. Find the least common denominator (LCD).
  2. Multiply every term by the LCD to clear fractions.
  3. Solve the resulting equation.
  4. Reject extraneous solutions that make a denominator zero.
Always check: clearing denominators can create solutions that don't work in the original.
Solving rational equations Solving rational equations Solving rational equations 1. find the LCD 2. multiply every term by the LCD 3. solve the result 4. reject extraneous solutions
Clear the fractions, solve, then check.
Extraneous check Extraneous check Extraneous check a solution that makes any denominator zero must be rejected
Rejecting an extraneous solution.

Clear the denominators:

\[\text{multiply every term by the LCD}\]
multiply every term by the least common denominator to clear fractions
A solution that makes a denominator \(0\) is extraneous.

How to solve

  1. Factor denominators and find the LCD.
  2. Multiply every term by the LCD.
  3. Solve the resulting polynomial equation.
  4. Check each solution against the original denominators.
Example 1 β€” Proportion
Solve \(\dfrac{1}{x-1}=\dfrac{2}{x+1}\).
Solution

Cross-multiply.

\(x+1\)\(=\)\(2(x-1)\)
\(x+1\)\(=\)\(2x-2\)
\(x\)\(=\)\(3\)
x equals 3
Example 2 β€” Extraneous solution
Solve \(\dfrac{x}{x-3}=\dfrac{3}{x-3}+2\).
Solution

Multiply by \((x-3)\).

\(x\)\(=\)\(3+2(x-3)\)
\(x\)\(=\)\(2x-3\)
\(x\)\(=\)\(3\)

But \(x=3\) makes a denominator zero β€” no solution.

x equals 3 is extraneous, so there is no solution
Example 3 β€” Clear the fractions
Solve \(\dfrac{2}{x}+\dfrac13=1\).
Solution

Multiply every term by \(3x\).

\(6+x\)\(=\)\(3x\)
\(6\)\(=\)\(2x\)
\(x\)\(=\)\(3\)
x equals 3
Example 4 β€” Why check
Why must you check solutions of rational equations?
Solution

Multiplying by a variable expression can introduce values that make a denominator zero β€” extraneous solutions that must be rejected.

because clearing denominators can create extraneous solutions

Common pitfalls

Multiply every term, not just some, by the LCD.
Always check for extraneous solutions.
A value that zeros a denominator is rejected, even if the algebra gives it.

Frequently asked questions

How do you solve a rational equation?

Multiply every term by the LCD to clear fractions, then solve.

What is an extraneous solution?

A value the algebra produces that makes a denominator zero and must be rejected.

Why do extraneous solutions appear?

Multiplying by a variable expression can introduce invalid values.

Do you always need to check?

Yes β€” check every solution in the original equation.