Algebraic fraction products and domains

A quotient has a domain before simplification

For 6x/(9x²), the denominator requires x ≠ 0. Divide numerator and denominator by 3x to obtain 2/(3x), still with x ≠ 0. Cancellation divides by a common non-zero factor.

For (x + 3)/x, you cannot cancel the x across an addition. It equals 1 + 3/x, not 3, and still requires x ≠ 0. Testing x = 2 exposes the incorrect cancellation.

Multiply factors across the product

Cancel the common factor 3 and one factor of b. The two factors of a multiply to a²; they do not cancel each other because they are both in numerators.

Division needs a non-zero divisor too

The second fraction must not be zero because it is the divisor. Taking its reciprocal is valid only under that condition. State restrictions from the original expression before simplifying; a cancelled factor does not restore an originally excluded value.

Check your understanding 1 An apparently harmless simplified result

Simplify 2x/3÷4x/9 and state the restriction.

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Multiply by the reciprocal: 2x/3 × 9/(4x) = 3/2 for x ≠ 0. At zero the original divisor is zero, so the constant simplified expression does not extend the original domain.

Use a numerical check

Check the result for x = 2, then explain why testing a few values cannot replace an algebraic justification.