Solving Rational Equations

A rational equation is solved by clearing the denominators — multiply every term by the LCD so the fractions vanish — but that step isn't reversible, so every candidate must be checked against the equation's domain restrictions before you trust it.

By the end you'll be able to clear denominators to solve a rational equation, and reject any extraneous solution that turns out to be excluded from the domain.

Predict: drag c until it matches a — does the candidate solution survive the domain check, or get struck as extraneous? Drag the sliders below to check.

This solves \(\dfrac{x}{x-a} = \dfrac{c}{x-a} + 2\). Clearing the denominator \((x-a)\) always gives a candidate — but the restriction x = a (open circle) is off-limits from the start. Watch the candidate (filled circle) on the number line: when it lands on the restriction, it gets struck as extraneous.

\(\dfrac{x}{x-3} = \dfrac{1}{x-3} + 2\)

Clear \((x-a)\): \(x = c + 2(x-a)\) → \(x = 2a - c = \)
Restriction: \(x \ne \)

Valid solution: x = 5.

Number line — restriction (open circle) vs. candidate solution (filled circle)
restriction x = a (excluded) candidate solution

Clearing denominators turns a rational equation into an ordinary polynomial one — but the multiplication step can manufacture roots the original equation never had, so checking every candidate is not optional.

Procedural

Solve a rational equation in three steps:

  • Find the restrictions. Set every denominator equal to zero — those x-values are excluded from the start.
  • Clear the LCD. Multiply every term on both sides by the least common denominator so no fractions remain, then solve the resulting polynomial equation.
  • Check each candidate. Compare every candidate solution to the restrictions from step one — a candidate that matches a restriction is extraneous and must be rejected.
Formal

Multiplying both sides of an equation by an expression that can equal zero is not a reversible operation — it can introduce solutions that satisfy the cleared equation but not the original one. That's exactly what happens when the LCD is zero at some x: the cleared equation is happy with that x, but the original equation has a zero denominator there and is undefined. So the domain restrictions found before clearing remain in force afterward, and any candidate equal to one of them is rejected, no matter how the algebra worked out.

Applied

This shows up any time a rate or ratio problem is modeled with a variable in a denominator — for instance solving for a time or a rate at which two quantities balance. If the algebra produces an answer that would have made a rate zero in the original setup, that answer isn't physically meaningful, and the domain check catches it before it's reported as a result.

Worked example

Solve \(\dfrac{x}{x-3} = \dfrac{3}{x-3} + 2\). Restriction: \(x \ne 3\). Clear \((x-3)\): \(x = 3 + 2(x-3) = 2x - 3\). Solve: \(x - 2x = -3\), so \(-x = -3\), giving \(x = 3\). Check: \(x = 3\) matches the restriction, so it's extraneous — the equation has no solution.

Your turn

Solve \(\dfrac{x}{x-1} = \dfrac{4}{x-1} + 3\). Restriction: \(x \ne 1\). Clear \((x-1)\): \(x = 4 + 3(x-1) = 3x + 1\). Solve: \(x - 3x = 1\), so \(-2x = 1\), giving \(x = \) ____. Now check it against the restriction: is it extraneous, or valid?

Reveal the answer

\(-2x = 1\) gives \(x = -1/2\). Since \(-1/2 \ne 1\), the candidate doesn't hit the restriction, so it's valid: \(x = -1/2\) is the solution. Set a = 1 and c = 4 on the sliders above to see an analogous case (the demo uses a fixed "+2" instead of "+3" — the same idea still applies: the candidate only becomes extraneous when it lands on the restriction).

More info — why clearing denominators isn't reversible

Multiplying both sides of an equation by a nonzero number always preserves the solution set — you can divide back to undo it. But multiplying by an expression like \((x-a)\) is different: at \(x = a\), you're multiplying both sides by zero, and \(0 = 0\) is true no matter what the original equation said. That's why the cleared equation can "agree" at a value where the original one is undefined. The fix isn't a different algebra trick — it's simply checking every candidate against the restrictions you identified before you started clearing. See the Purplemath link in Dive deeper for more worked examples of this check.

Check your understanding

Question 1 of 4

Solve \(\dfrac{x}{x-5} = \dfrac{5}{x-5} + 2\). Clear the denominator, solve, and check the candidate against the restriction.

Question 2 of 4

After clearing denominators in a rational equation, a candidate solution turns out to equal a value you'd already excluded from the domain. What do you conclude?

Question 3 of 4

Before solving \(\dfrac{x}{x-2} + \dfrac{1}{x+3} = 0\), which values must every candidate solution be checked against?

Question 4 of 4

Solve \(\dfrac{x}{x+2} = \dfrac{5}{x+2} - 1\), and check the candidate against the restriction.

Recap

  • Find the domain restrictions first: set every denominator equal to zero — those x-values are excluded before you solve anything.
  • Clear the LCD by multiplying every term on both sides, then solve the resulting polynomial equation.
  • Multiplying by an expression that can be zero isn't reversible, so it can introduce extraneous solutions.
  • Check every candidate against the restrictions: a candidate equal to a restriction is extraneous and must be rejected, even if the cleared equation "solved cleanly."

Dive deeper

Sources

  • Solving Rational Equations