Guides / Rational expressions

Rational equations: clear the denominators, then check for extraneous solutions

An x in a denominator means some values of x are off-limits before you start. Multiply through by the LCD and the fractions disappear, but the off-limits values don't, and the points go when an answer on that list is kept.

Why it works

Multiplying both sides of an equation by the same nonzero number keeps it true, which is the same idea as clearing numeric denominators. The difference here is that the LCD has x in it. At the excluded values the LCD is zero, and multiplying by zero turns any equation into 0 = 0. So the cleared equation can have answers the original equation never had. They don't come from an arithmetic slip; they come from a step that is legal everywhere except at those values.

That makes the check easy. You don't need to substitute every answer (though it never hurts). You only need to compare each answer with the list you wrote at the start. This is a different source of extraneous solutions from squaring both sides, where the list isn't known in advance and substitution is the only check.

Example 1: one variable denominator

Solve 2x + 13 = 1.

1. x ≠ 0The only denominator with x in it is x itself. Write the exclusion before solving.
2. 3x · 2x + 3x · 13 = 3x · 1The LCD of x and 3 is 3x. Every term gets it, including the 1 on the right.
3. 6 + x = 3xCancel: 3x/x = 3, so 3 · 2 = 6; 3x/3 = x; and 3x · 1 = 3x.
4. x = 3Subtract x, divide by 2. 3 isn't on the excluded list. Check: 2/3 + 1/3 = 1.

Example 2: the only answer is extraneous

Solve xx − 3 = 3x − 3 + 2.

1. x ≠ 3x − 3 = 0 at x = 3.
2. x = 3 + 2(x − 3)Multiply every term by (x − 3). The 2 becomes 2(x − 3).
3. x = 2x − 3Distribute: 3 + 2x − 6.
4. x = 3Subtract 2x, then multiply by −1.
5. No solutionThe only candidate is on the excluded list. Substituting gives 3/0 on both sides, which is undefined.

Example 3: factor a denominator to find the LCD

Solve xx − 2 + 1x + 1 = 6x² − x − 2.

1. x² − x − 2 = (x − 2)(x + 1)Factor the quadratic denominator. It's the product of the other two, so the LCD is (x − 2)(x + 1).
2. x ≠ 2, x ≠ −1Both factors appear as denominators.
3. x(x + 1) + 1(x − 2) = 6Multiply every term by (x − 2)(x + 1). Each fraction keeps the factor its denominator lacked.
4. x² + 2x − 8 = 0Expand: x² + x + x − 2 = 6, then subtract 6.
5. (x + 4)(x − 2) = 0Factor: 4 and −2 multiply to −8 and add to 2.
6. x = −4 or x = 2Two candidates from the zero-product property.
7. x = −4 only2 is excluded. Check −4: −4/(−6) + 1/(−3) = 2/3 − 1/3 = 1/3, and 6/(16 + 4 − 2) = 6/18 = 1/3.

Common mistakes, and the exact line they happen on

1. Not multiplying the term that has no fraction

Solve 2x + 13 = 1 (Example 1).

1. 6 + x = 1This is the mistake. The two fractions were multiplied by 3x and the 1 on the right was left alone. This gives x = −5, and the check fails: 2/(−5) + 1/3 = −1/15.

The LCD multiplies the whole side, and a whole number is part of the side. It's the same slip as in equations with numeric fractions: the term without a denominator is the one that gets forgotten.

2. Cross-multiplying across a plus sign

Solve 1x + 12 = 13.

1. x + 2 = 3This is the mistake. Cross-multiplying only works for one fraction equal to one fraction. Here the left side is a sum, and "flipping the denominators up" has no meaning. x = 1 fails: 1 + 1/2 = 3/2, not 1/3.
2. 6 + 3x = 2xCorrect line: multiply every term by the LCD 6x.
3. x = −6Subtract 3x. Check: −1/6 + 1/2 = −1/6 + 3/6 = 2/6 = 1/3.

3. Keeping the extraneous answer

In Example 3 the algebra ends with x = −4 or x = 2, and writing both as the final answer loses the point even though every line was right. Substitute 2: the first fraction is 2/0. The cleared equation on line 3 is perfectly happy with x = 2, which is exactly why you can't check in the cleared equation. Check against the list, or in the original.

4. Throwing out an answer that isn't on the list

Solve xx + 2 + 1 = 2x + 2.

1. x ≠ −2The only excluded value.
2. x + (x + 2) = 2Multiply every term by (x + 2).
3. x = 02x + 2 = 2, so 2x = 0.
4. No solutionThis is the mistake. Zero "looks like" a value that breaks fractions, so it got rejected. But 0 isn't on the list: the denominator is x + 2, which is 2 at x = 0. Check: 0/2 + 1 = 1 and 2/2 = 1. The answer is x = 0.

An answer is extraneous only if it makes a denominator of the original equation zero. Nothing else disqualifies it, and zero is as good an answer as any other number.

Practice

1. Solve 5x − 12 = 3x.

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   x ≠ 0. LCD 2x: 10 − x = 6, so x = 4. Check: 5/4 − 1/2 = 3/4 ✓.

2. Solve xx − 4 = 4x − 4 + 2.

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   x ≠ 4. Multiply by (x − 4): x = 4 + 2x − 8, so x = 4. That's excluded, so there is no solution.

3. Solve 3x + 1 + 12 = 1.

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   x ≠ −1. LCD 2(x + 1): 6 + (x + 1) = 2(x + 1), so x + 7 = 2x + 2 and x = 5. Check: 3/6 + 1/2 = 1 ✓.

4. Solve 2x − 1 + 1x + 1 = 6x² − 1.

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   x² − 1 = (x − 1)(x + 1), so x ≠ ±1. Clear: 2(x + 1) + (x − 1) = 6, 3x + 1 = 6, x = 5/3. Not excluded. Check: both sides equal 27/8 ✓.

5. Solve xx − 3 + 2x + 2 = 15x² − x − 6.

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   x² − x − 6 = (x − 3)(x + 2), so x ≠ 3, −2. Clear: x(x + 2) + 2(x − 3) = 15, x² + 4x − 21 = 0, (x + 7)(x − 3) = 0. x = 3 is extraneous. Only x = −7. Check: 7/10 − 2/5 = 3/10 and 15/50 = 3/10 ✓.