Formulae and restrictions

A formula relates several quantities

The formula v = u + at relates final velocity, initial velocity, acceleration and time. Making t the subject means expressing it in terms of the other quantities.

Subtract u from both sides: v-u = at. If a ≠ 0, divide by a:

If a = 0, that division is undefined. The original formula instead says v = u and does not determine t. A rearranged formula must retain the condition that made its operations valid.

To make x the subject of y = (2x-3)/5, multiply both sides by five, then add three, then divide by two:

Here the divisors five and two are always nonzero. Check by replacing x in the original formula: [2(5y + 3)/2-3]/5 = y.

A denominator restricts the original input

For (x²-9)/(x-3), the input x = 3 is excluded because the denominator would be zero. Factor the numerator before cancelling:

The simplified formula has a value at three, but the original expression does not. Keep the original restriction. Cancelling does not fill the missing input.

At x = 4, both forms give seven. At x = 3, only the unrestricted expression x + 3 gives six; it is not a value of the original fraction.

Check your understanding 1 A tempting cancellation

A learner writes (x + 6)/x = 6. Explain why cancelling the x is invalid, give a counterexample and write a valid form with its restriction.

Your counterexample, corrected expression and restriction

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The numerator is a sum, not x times six. At x = 2, the original is 8/2 = 4, not six. Splitting the division gives 1 + 6/x, for x ≠ 0. Only a nonzero factor of the entire numerator and denominator can be cancelled. Now compare 6x/x = 6 for x ≠ 0: there x is a factor of the whole numerator.

Combine fractions with a common denominator

For 2/x + 1/(x + 1), the original restrictions are x ≠ 0 and x ≠ -1. Use the product x(x + 1) as a common denominator:

Each numerator was multiplied by the same factor as its own denominator. The restrictions remain. At x = 1, the original gives 2 + 1/2 = 5/2, matching the combined expression.

If the task asks for one fraction, the uncombined sum has the right value but does not yet meet the requested form. If it asks for domain restrictions, a final expression alone does not supply them.

Zero on the right changes what can be inferred

Solve (x²-9)/(x-3) = 0. Start with x ≠ 3. On that domain the equation is equivalent to x + 3 = 0, giving x = -3. Check in the original: the numerator is zero and the denominator is -6, so the fraction equals zero.

Setting only the original numerator to zero gives candidates x = 3 and x = -3. The first must be rejected because it is outside the original domain. A zero numerator with a zero denominator is undefined, not zero.

Check your understanding 2 Simplify, then test a proposed solution

Simplify (x²-4)/(x + 2) with its original restriction. A learner uses the simplified form to claim that the original fraction equals -4 at x = -2. Is that valid?

Simplification, restriction and judgement

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The numerator is (x-2)(x + 2), so the fraction equals x-2 for x ≠ -2. The proposed input is excluded: the original fraction becomes 0/0, which is undefined. Although x-2 has value -4 there, that extended value does not belong to the original expression.

Before you move on

For a rational expression, write the original restrictions before simplifying. For an equation, substitute candidate solutions into the original statement. These checks protect different things: allowed inputs, equivalent forms and actual solutions.