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MathChapter 14: Rational Expressions

Notes

Solving Rational Equations Notes

Use domain restrictions and the LCD to solve rational equations, then reject denominator-zero or otherwise extraneous candidates.

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A rational equation contains at least one rational expression. Clearing denominators can turn it into a familiar equation, but it can also produce candidates the original equation never allowed.

The domain-first method

Solve rational equations safelyA domain-first workflow prevents denominator-zero candidates from being accepted.
  1. \(\text{denominators}\ne0\)

    List excluded values before changing the equation.

  2. \(\operatorname{LCD}\)

    Include every distinct factor at its greatest power.

  3. \(\operatorname{LCD}(\text{every term})\)

    Distribute the LCD to every term on both sides.

  4. \(\text{candidate values}\)

    Solve the resulting equation.

  5. \(\text{valid solution?}\)

    Reject excluded values and verify the original equation.

Worked example

Clear denominators and check

Solve \(\frac{2}{x}+1=\frac{3}{x}\).

  1. Restrict

    \(x\ne0\).

  2. Clear

    Multiply every term by \(x\): \(2+x=3\).

  3. Solve and check

    \(x=1\), which is allowed and makes both sides equal \(3\).

\(x=1\).

When cross multiplication is valid

Single proportion versus a sum

Valid

For \(\frac{A}{B}=\frac{C}{D}\), with \(B,D\ne0\), cross multiplication gives \(AD=BC\).

Not directly valid

For \(\frac{A}{B}+E=\frac{C}{D}\), first use the LCD or combine one side into a single fraction.

Why

Cross multiplication is shorthand for multiplying both sides of one-fraction-equals-one-fraction by \(BD\).

Worked example

Use a proportion

Solve \(\frac{x+2}{x-1}=\frac43\).

  1. Restriction: \(x\ne1\).

  2. Cross multiply: \(3(x+2)=4(x-1)\).

  3. Solve: \(3x+6=4x-4\), so \(x=10\).

  4. Substitution gives \(12/9=4/3\), so the candidate is valid.

\(x=10\).

Extraneous solutions and no-solution equations

Extraneous solution
A value produced by transformed algebra that does not satisfy the original equation or makes an original denominator zero.
Worked example

A candidate that must be rejected

Solve \(\frac{x}{x-2}=\frac{2}{x-2}\).

  1. Restrict

    \(x\ne2\).

  2. Clear

    Multiplying by \(x-2\) produces \(x=2\).

  3. Reject

    The only candidate makes the original denominator zero.

No solution.
Candidate-check table
CandidateOriginal denominators nonzero?Original equality true?Decision
\(x=2\) aboveNoUndefinedReject
\(x=10\) in the proportionYesYesAccept
Any allowed candidate that fails equalityYesNoReject

SAT strategy

Common mistakes and traps

  • Cross multiplying an equation that is not one fraction equal to one fraction.
  • Forgetting to distribute the LCD to a constant or signed term.
  • Canceling denominators across addition.
  • Accepting a value that zeros an original denominator.
  • Checking only the transformed equation.
Mini check

Domain check

If the algebra from \(\frac{x+1}{x-3}=\frac4{x-3}\) produces \(x=3\), what is the solution set?

Show answer and explanation

Answer: No solution.

The candidate \(3\) is excluded by the original denominator.

Key takeaways

Key takeaways

  • Restrictions come first.
  • The LCD must multiply every term.
  • Cross multiplication is a limited shortcut.
  • Every candidate must be verified in the original equation.
  • An excluded only candidate means no solution.
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Put these notes into practice

Apply the ideas with SAT-style questions, then reinforce key details with flashcards.