What Is a Rational Equation?
A rational equation is any equation that contains at least one rational expression — a fraction with a polynomial in the numerator, the denominator, or both. Examples include 1/x + 1/3 = 1/2, (x + 1)/(x − 2) = 3, and 2/(x + 1) + 3/(x − 1) = 5/(x² − 1).
x that makes a denominator equal to zero is excluded from the domain. For instance, in (x + 1)/(x − 2) = 3, x = 2 is not allowed because it would create division by zero.Before solving any rational equation, you should identify these restricted values by setting each denominator equal to zero and solving. Write them down — you will need to check your final answers against this list.
The standard technique for solving rational equations is to find the least common denominator (LCD) of all fractions in the equation and multiply every term by it. This clears all the fractions and transforms the rational equation into a simpler polynomial equation that you can solve with familiar methods like factoring or the quadratic formula.
Finding the LCD
How to find it
For simple numeric denominators: Find the LCM. For example, if the denominators are 3, 4, and 6, the LCD is 12.
For polynomial denominators: Factor each denominator completely, then take each unique factor to its highest power.
Denominators are x and x + 2. These share no common factors, so LCD = x(x + 2).
Denominators are x − 3, x + 3, and x² − 9. Factor the last one: x² − 9 = (x − 3)(x + 3). The unique factors are (x − 3) and (x + 3), so LCD = (x − 3)(x + 3). Notice that x² − 9 didn't add any new factors — it was already the product of the other two denominators.
Denominators are x² and x³. The LCD is x³ (take x to its highest power).
Do not just multiply all the denominators together. While that works, it often creates unnecessarily large expressions. Finding the true LCD keeps the algebra manageable.
Multiplying Through to Clear Fractions
Solve 1/x + 1/3 = 1/2. The denominators are x, 3, and 2, so LCD = 6x.
Multiply every term by 6x: 6x · (1/x) + 6x · (1/3) = 6x · (1/2).
Simplify: 6 + 2x = 3x.
Solve: 6 = x.
Check: 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2. Valid.
x = 6Solve (x + 1)/(x − 2) = 3. The only denominator is x − 2, so multiply both sides by (x − 2):
x + 1 = 3(x − 2).
Expand: x + 1 = 3x − 6.
Solve: 7 = 2x, so x = 7/2.
Check: (7/2 + 1)/(7/2 − 2) = (9/2)/(3/2) = 3. Valid.
x = 7/2You must multiply every term, including terms that are not fractions. If you have x + 2/x = 3, the x term must also be multiplied by the LCD: x · x + x · (2/x) = x · 3 gives x² + 2 = 3x.
Checking for Extraneous Solutions
When you multiply both sides of an equation by an expression containing a variable, you may introduce extraneous solutions — values that satisfy the transformed equation but make a denominator zero in the original equation. You must check every solution.
Solve x/(x − 3) − 3/(x − 3) = 1. LCD = x − 3.
Multiply through: x − 3 = x − 3.
This simplifies to 0 = 0, which is always true. Does that mean every number is a solution? No — x = 3 is restricted because it makes the denominator zero.
all real numbers except x = 3Solve 2/(x − 1) = (x + 1)/(x − 1). LCD = x − 1.
Multiply through: 2 = x + 1, so x = 1.
But x = 1 makes the denominator x − 1 = 0! This is an extraneous solution.
no solutionExtraneous solutions are not errors in your algebra — they are a natural consequence of the clearing process. The key is to always verify your answers at the end.
Worked Examples
Solve 3/(x + 2) + 2/(x − 2) = 12/(x² − 4).
Factor: x² − 4 = (x + 2)(x − 2). LCD = (x + 2)(x − 2). Restrictions: x ≠ 2, x ≠ −2.
Multiply through: 3(x − 2) + 2(x + 2) = 12.
Expand: 3x − 6 + 2x + 4 = 12.
Simplify: 5x − 2 = 12, so 5x = 14 and x = 14/5.
Check restrictions: 14/5 ≠ 2 and 14/5 ≠ −2. Valid.
x = 14/5Solve x/(x − 4) + 1 = 4/(x − 4).
LCD = x − 4. Restriction: x ≠ 4.
Multiply: x + (x − 4) = 4.
Simplify: 2x − 4 = 4, so x = 4.
But x = 4 is restricted — extraneous!
No solution.Solve 5/(x + 3) = 3/(x − 1).
LCD = (x + 3)(x − 1). Cross-multiply: 5(x − 1) = 3(x + 3).
Expand: 5x − 5 = 3x + 9.
Solve: 2x = 14, x = 7.
Check: 5/10 = 1/2 and 3/6 = 1/2. Valid.
x = 7Solve 1/(x − 2) + 1/(x + 2) = 4/(x² − 4).
LCD = (x − 2)(x + 2).
Multiply: (x + 2) + (x − 2) = 4.
Simplify: 2x = 4, x = 2.
But x = 2 is restricted.
No solution.Common Mistakes
This is the number one error in rational equations. Every time you clear fractions by multiplying by a variable expression, you must check your answers against the restricted values. Skipping this step will cost you points on every exam.
If one side of the equation has a term without a fraction (like + 3), you still must multiply it by the LCD. Students often clear the fractions but leave constants untouched, producing wrong answers.
When a denominator is a sum or difference (like x + 3), it is already fully factored. Do not break it into x and 3 as separate factors. The LCD of x and x + 3 is x(x + 3), not 3x.
After multiplying by the LCD, you often need to distribute. Watch your signs carefully — a negative sign in front of a fraction applies to the entire numerator, not just the first term. For instance, −(x + 1) is −x − 1, not −x + 1.
Write the restricted values at the very top of your work before you start solving. This way you cannot forget to check them.
If you want to verify your work on any rational equation, scan it with Solver AI for a complete step-by-step solution.