Fractions with letters
Tonight's idea
Put a number in for \(x\) and a fraction with letters becomes an ordinary fraction. So it must obey the ordinary rules.
Cancelling. \(\tfrac69 = \tfrac{2 \times 3}{3 \times 3} = \tfrac23\) because \(\tfrac33 = 1\), and multiplying by 1 changes nothing. What cancels is a common factor: something that multiplies the whole top and the whole bottom. So factor first, then cancel.
Never cancel across a plus sign. In \(\dfrac{x + 2}{x + 5}\) the \(x\) is added on, not multiplied. It is a term, not a factor. Try \(x = 1\): the fraction is \(\tfrac36 = \tfrac12\), but crossing out the \(x\)'s gives \(\tfrac25\).
Adding. Fractions add only when the bottoms match. Multiply each one by 1, chosen to give both the same bottom: \(\dfrac1a + \dfrac1b = \dfrac{b}{ab} + \dfrac{a}{ab} = \dfrac{a + b}{ab}\).
Factor, then cancel factors. To test any step, put a number in before and after. If the two values differ, the step was wrong.
Watch one: simplify, then add
Simplify \(\dfrac{x^2 + 3x}{x^2 + 5x + 6}\). Factor the top and the bottom (the bottom by the sum and product hunt):
\[\frac{x^2 + 3x}{x^2 + 5x + 6} = \frac{x(x + 3)}{(x + 2)(x + 3)} = \frac{x}{x + 2}.\]Test with \(x = 1\): before, \(\tfrac{4}{12} = \tfrac13\); after, \(\tfrac13\).
Add \(\dfrac2x + \dfrac{3}{x + 1}\). The common bottom is \(x(x + 1)\):
\[\frac{2(x + 1)}{x(x + 1)} + \frac{3x}{x(x + 1)} = \frac{5x + 2}{x(x + 1)}.\]Test with \(x = 1\): \(2 + 1.5 = 3.5\), and \(\tfrac72 = 3.5\).
Your turn
- Simplify: (a) \(\dfrac{12x^2}{8x}\); (b) \(\dfrac{4x + 8}{x + 2}\); (c) \(\dfrac{x^2 - 9}{x + 3}\).
- Factor, then cancel: (a) \(\dfrac{x^2 + 7x + 12}{x^2 + 3x}\); (b) \(\dfrac{x^2 - 4}{x^2 + x - 6}\). Test each answer with \(x = 1\).
- Write as a single fraction: (a) \(\dfrac x2 + \dfrac x3\); (b) \(\dfrac1x + \dfrac{1}{x + 1}\); (c) \(\dfrac{3}{x - 1} - \dfrac{2}{x + 1}\).
- Sam writes \(\dfrac{x + 6}{x + 2} = 3\), saying "the \(x\)'s cancel and \(6 \div 2 = 3\)". (a) Test Sam's claim with \(x = 2\). (b) Solve \(\dfrac{x + 6}{x + 2} = 3\) properly. For how many values of \(x\) is Sam right? (c) Say in one sentence what Sam did wrong.
Stretch: a fraction inside a fraction
Simplify \(\dfrac{1 + \frac1x}{1 - \frac{1}{x^2}}\) to one fraction with no fractions inside it. Test your answer with \(x = 2\). Which three values of \(x\) are not allowed in the original, and why?
Answers
- (a) \(\dfrac{3x}{2}\): cancel the common factor \(4x\). (b) \(\dfrac{4(x + 2)}{x + 2} = 4\). (c) \(\dfrac{(x - 3)(x + 3)}{x + 3} = x - 3\), using the difference of two squares.
- (a) \(\dfrac{(x + 3)(x + 4)}{x(x + 3)} = \dfrac{x + 4}{x}\). At \(x = 1\): \(\tfrac{20}{4} = 5\) and \(\tfrac51 = 5\). (b) \(\dfrac{(x - 2)(x + 2)}{(x - 2)(x + 3)} = \dfrac{x + 2}{x + 3}\). At \(x = 1\): \(\tfrac{-3}{-4} = \tfrac34\) and \(\tfrac34\).
- (a) \(\dfrac{3x}{6} + \dfrac{2x}{6} = \dfrac{5x}{6}\). (b) \(\dfrac{(x + 1) + x}{x(x + 1)} = \dfrac{2x + 1}{x(x + 1)}\). (c) \(\dfrac{3(x + 1) - 2(x - 1)}{(x - 1)(x + 1)} = \dfrac{x + 5}{(x - 1)(x + 1)}\). Watch the sign: \(-2(x - 1) = -2x + 2\).
- (a) \(\tfrac{8}{4} = 2\), not 3. (b) Multiply both sides by \(x + 2\): \(x + 6 = 3x + 6\), so \(2x = 0\) and \(x = 0\). Sam is right for exactly one value, \(x = 0\), where the fraction is \(\tfrac62\). (c) The \(x\) is a term added to the top and bottom, not a factor of them, so it cannot be cancelled.
Stretch. The top is \(\dfrac{x + 1}{x}\) and the bottom is \(\dfrac{x^2 - 1}{x^2} = \dfrac{(x - 1)(x + 1)}{x^2}\). Dividing is multiplying by the flipped fraction: \(\dfrac{x + 1}{x} \times \dfrac{x^2}{(x - 1)(x + 1)} = \dfrac{x}{x - 1}\). At \(x = 2\): \(\tfrac{1.5}{0.75} = 2\) and \(\tfrac21 = 2\). Not allowed: \(x = 0\) (it makes \(\tfrac1x\) meaningless), and \(x = 1\) and \(x = -1\) (they make the bottom zero). The simplified form hides two of the three, so note them before you cancel.