Algebraic fraction equations
Objective. Solve rational equations and reject invalid values.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Algebraic fraction equations are solved by multiplying every term by the lowest common denominator to clear all fractions, turning the equation into a polynomial equation that can be solved normally, then checking that no solution makes an original denominator zero. After clearing denominators, expand brackets, collect like terms, and solve the resulting linear or quadratic equation. Any solution that would make a denominator in the original equation equal to zero must be rejected as invalid, since division by zero is undefined. For example, solving 3/(x−1) = 2/x by cross-multiplying gives 3x = 2(x−1), so 3x=2x−2, giving x=−2, which is valid since neither denominator is zero.
Worked example
Solve the equation 2/(x+1) + 1/x = 1, rejecting any invalid solutions.
- 1Multiply every term by x(x+1): 2x + (x+1) = x(x+1)
- 2Simplify: 3x+1 = x²+x
- 3Rearrange: x²−2x−1=0, giving x = (2±√(4+4))/2 = 1±√2
- 4Check neither solution makes x=0 or x=−1, so both are valid
Answer: x = 1+√2 or x = 1−√2
Concept mastery
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Practice
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Start practisingCommon mistake. Students frequently forget to check for and reject extraneous solutions that make a denominator zero, accepting x=1 as a solution to an equation containing 1/(x−1) even though it makes that fraction undefined; always substitute solutions back to check.
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