Skip to main content

Topic

Solving rational equations

My Progress

Pug Score

0%

Best Streak

0 in a row

Study Points

+0

Unlock Full Access
Adaptive practice
Video lessons

Read

Solving Rational Equations

A rational equation contains at least one algebraic fraction with the variable in a denominator. Solve one by noting the domain restrictions, finding the LCD, clearing the fractions, solving the resulting equation, and checking the solution isn't a restricted value, with a full worked example.

What a rational equation is

A rational equation is an equation containing at least one algebraic fraction, with the variable somewhere in a denominator. Solving one means clearing the fractions first, then solving what's left as an ordinary equation.

Solving a rational equation Solve 3/(x+1) = 2/(x-1). Step 1: the LCD is (x+1)(x-1). Step 2: cross-multiply to clear fractions: 3(x-1) = 2(x+1). Step 3: solve: 3x-3 = 2x+2, so x=5. Step 4: check x=5 is not a restricted value (x cannot be -1 or 1) -- it is valid. 1 3/(x+1) = 2/(x−1) restrictions: x ≠ −1, 1 2 clear fractions (cross-multiply): 3(x−1) = 2(x+1) 3 3x − 3 = 2x + 2 → x = 5 4 check: x = 5 is not a restricted value 3/6 = 2/4 = 0.5 ∞ x = 5 is the valid solution
Clear the fractions using the LCD, solve, then check against the restrictions.
  1. Note the restrictions: any value that makes a denominator zero is not allowed in the answer.
  2. Find the LCD (least common denominator) of all the fractions.
  3. Clear the fractions by multiplying every term by the LCD (or cross-multiplying, for a single fraction on each side).
  4. Solve the resulting equation.
  5. Check that the solution isn't a restricted value.

Solve 3/(x+1) = 2/(x−1). The restrictions are x ≠ −1 and x ≠ 1. Cross-multiplying clears the fractions: 3(x−1) = 2(x+1), which expands to 3x − 3 = 2x + 2, giving x = 5. Since 5 is not a restricted value, it's valid — and checking confirms 3/6 = 2/4 = 0.5.

Occasionally, solving the cleared equation produces a value that turns out to equal one of the restricted values. When that happens, it must be rejected — not because the algebra was wrong, but because that value was never in the domain to begin with. This is a different issue from simplifying complex fractions, but both require tracking domain restrictions carefully. For fraction arithmetic building blocks, see adding and subtracting rational expressions.

Related lessons