Algebraic fractions calculator

Numeric fractions are forgiving. Get 3/4 + 1/2 slightly wrong and you'll probably notice, because the answer is a number you can sanity-check in your head. Algebraic fractions don't give you that safety net. The "wrong" answer still looks like a perfectly reasonable expression, which is exactly why small mistakes in this topic tend to survive all the way to the final exam answer.

The Rule That Doesn't Change From Arithmetic

Adding or subtracting algebraic fractions still needs a common denominator, same as 1/3 + 1/4 does. The difference is that your denominators are now expressions like x+2 or 3x−1, so the "common denominator" is the product of both expressions rather than a small whole number.

Take (x+1)/x + 3/(x+2). The common denominator is x(x+2). Multiply the first fraction's numerator and denominator by (x+2), and the second by x: you get [(x+1)(x+2)]/[x(x+2)] + [3x]/[x(x+2)]. Expand the top: (x+1)(x+2) = x² + 3x + 2. Add the second numerator: x² + 3x + 2 + 3x = x² + 6x + 2. Final answer: (x² + 6x + 2) / [x(x+2)].

Multiplication and Division Skip the Common Denominator Step Entirely

This is where students often overcomplicate things. Multiplying algebraic fractions doesn't need a common denominator at all — multiply numerators together, multiply denominators together, done. (x+3)/2 × x/(x−1) becomes [x(x+3)] / [2(x−1)] directly, no extra steps.

Division is multiplication by the reciprocal. (x+3)/2 ÷ x/(x−1) flips the second fraction and multiplies: (x+3)/2 × (x−1)/x = [(x+3)(x−1)] / (2x). Forgetting to flip is the single most common division mistake in this topic, full stop.

Simplifying After You've Combined

A combined fraction isn't finished until you've factored both top and bottom and cancelled anything that appears in both. Take (x² − 9) / (x² + 3x). Factor the numerator: (x−3)(x+3). Factor the denominator: x(x+3). The (x+3) term appears in both, so it cancels, leaving (x−3)/x.

That cancellation step only works on factors that multiply the whole expression, never on terms you'd have to add or subtract apart. (x+3)/(x+5) cannot be simplified by "cancelling the x" — there's no shared multiplicative factor there, just two terms that happen to share a variable.

Restricted Values — The Step Most Students Skip

Every algebraic fraction has values of the variable that make a denominator zero, and those values are excluded from the domain. For (x−3)/[x(x+2)], x cannot equal 0 or −2, because either would divide by zero. State these restrictions even after simplifying, since cancelling (x+3) earlier doesn't undo the fact that x=−3 was originally excluded too — that restriction carries through the whole problem.

A Worked Exam-Style Example

Simplify: 2/(x−1) − 1/(x+1). Common denominator: (x−1)(x+1), which expands to x²−1. Numerator: 2(x+1) − 1(x−1) = 2x+2−x+1 = x+3. Final answer: (x+3)/(x²−1), with restrictions x≠1 and x≠−1.

That x²−1 denominator is a difference of two squares, which is exactly the kind of pattern worth recognising on sight in this topic — it saves a step of expansion every single time it shows up, and it shows up constantly in algebraic fraction questions.

FAQs

How do you add algebraic fractions with different denominators?

Find a common denominator by multiplying the two denominators together (or their lowest common multiple if they share factors), rewrite both fractions over that common denominator, then add the numerators.

Do you need a common denominator to multiply algebraic fractions?

No. Multiplying algebraic fractions just requires multiplying the numerators together and the denominators together directly, with no common denominator step needed.

How do you divide algebraic fractions?

Flip the second fraction (take its reciprocal) and multiply instead of dividing. This turns every division problem into a multiplication problem.

What does it mean to simplify a rational expression?

It means factoring both the numerator and denominator fully, then cancelling any factor that appears in both. Only multiplicative factors can be cancelled, never individual terms within a sum.

Why do algebraic fractions have restricted values?

Because any value of the variable that makes a denominator equal to zero is undefined in mathematics, so those specific values must be excluded from the solution.

What's the LCD of two algebraic fractions?

The lowest common denominator is the smallest expression that both original denominators divide into evenly. If the denominators share no common factors, the LCD is simply their product.

Most mistakes in this topic happen at the cancelling stage, not the combining stage — people cancel terms that were added rather than multiplied, and the wrong answer slips through looking entirely plausible. Factor first, cancel only shared multiplicative factors, and state your restrictions; that sequence catches almost every error before it reaches the final line.