How to add algebraic fractions with different denominators

A denominator cannot be zero. For 1/x, exclude x = 0. For 1/(x + 1), exclude x = −1. Record those restrictions before changing the expression, because a factor that disappears during simplification does not make the original fraction defined at its excluded value.

Read the fraction bars carefully. The numerator of (x + 3)/(x − 2) is the whole expression x + 3. Copying it as x + 3/(x − 2) would divide only the 3 and give a different expression. On a single line, use brackets to show exactly what is above and below each fraction bar.

For 3/x + 2/(x + 1), the denominators are x and x + 1. A common denominator is x(x + 1). Multiply the numerator and denominator of the first fraction by x + 1, and those of the second by x. The expression becomes [3(x + 1) + 2x]/[x(x + 1)].

Simplify the numerator: 3x + 3 + 2x = 5x + 3. The answer is (5x + 3)/[x(x + 1)], with x ≠ 0 and x ≠ −1. You added the numerators after making the denominators equal. Adding the original denominators to get 2x + 1 would not preserve the fractions' values.

You do not always need to multiply the denominators together. For 1/(2x) + 1/(3x), use 6x. The first fraction needs a factor of 3 and the second a factor of 2, giving 3/(6x) + 2/(6x) = 5/(6x), with x ≠ 0. Using 6x² can still produce an equivalent expression for allowed values, but creates extra work.

Consider the practice expression (x + 1)/x − (x − 2)/(x + 1). The common denominator is x(x + 1), so the combined numerator is (x + 1)(x + 1) − x(x − 2). Keep that subtraction outside the second bracket until you expand.

The first product is x² + 2x + 1. The second is x² − 2x. Subtracting the second product gives x² + 2x + 1 − (x² − 2x) = x² + 2x + 1 − x² + 2x = 4x + 1. Therefore the result is (4x + 1)/[x(x + 1)], with x ≠ 0 and x ≠ −1.

The minus sign changes both terms in the second product. If your answer contains an unexpected x² term or loses the 4x, inspect that step before starting the whole calculation again.

In (5x + 3)/[x(x + 1)], x is not a factor of the whole numerator: the 3 has no factor of x. Crossing out an x in 5x and another in the denominator would change the expression. Cancellation applies to multiplication factors shared by the entire numerator and denominator.

A shared denominator factor can make a sum shorter. For 1/(x − 1) + 1/[(x − 1)(x + 1)], use (x − 1)(x + 1). Only the first fraction needs multiplying by x + 1, so the sum is [(x + 1) + 1]/[(x − 1)(x + 1)] = (x + 2)/[(x − 1)(x + 1)]. The restrictions are x ≠ 1 and x ≠ −1.

If the final numerator factorises, compare its factors with the denominator's factors. For example, (x² − 1)/(x − 1) = [(x − 1)(x + 1)]/(x − 1) = x + 1 for x ≠ 1. Keep that restriction even though the simplified expression x + 1 can be evaluated at 1.

Check the first example at x = 2. The original gives 3/2 + 2/3 = 9/6 + 4/6 = 13/6. The combined expression gives (5 × 2 + 3)/[2(2 + 1)] = 13/6. Matching values can expose a wrong sign or multiplier, although one successful substitution is not a proof that two expressions are identical.

For the subtraction example at x = 2, the original gives 3/2 − 0/3 = 3/2. The result gives (4 × 2 + 1)/[2(2 + 1)] = 9/6 = 3/2. Choose a value that is allowed in every original denominator.

Use this working record when practising: original denominators and excluded values; common denominator; missing multiplier for each fraction; combined numerator with brackets; expanded and collected numerator; any common factors cancelled; substitution check. If a step fails, those lines show whether the problem came from the denominator, expansion or cancellation.