Four fraction failures you’d be a fool not to fix

Good God, will you get your fractions sorted out? You're meant to have been on top of them since primary school, but you've been studiously avoiding them for the last half-decade, haven't you?

If you'd put half as much effort into learning four simple rules as you have into convincing yourself fractions are impossibly difficult, you'd have found that they're not impossibly difficult... you just need to learn four simple rules. Dolt.

Rule #1: Times across the Top

Timesing fractions is ludicrously easy. You times across the top to get the top; you times across the bottom to get the bottom. No funny business:
$\frac{3}{7} \times \frac{4}{5}=\frac{3 \times 4}{7 \times 5}=\frac{12}{35}$

Same thing with algebraic fractions:
$\frac{(x+2)}{(x-4)} \times \frac{(2x-3)}{(x+3)} = \frac{(x+2)(2x-3)}{(x-4)(x+3)}$

Rule #2: Divide with Impala

There's an old Australian swagman - or rather, koala - named Impala. Impala The Koala, used to live on the Great Dividing Range. Impala and I have a reciprocal agreement: because he likes dividing fractions and (being Australian) likes turning things upside down, he helps out whenever there are fractions to be divided. This is how it goes down: Impala takes the second fraction and turns it upside-down (which is the right way up to him). Then I times the first fraction by whatever Impala gives me back, like this:
$\frac{3}{7} \div \frac{4}{5} = \frac{3}{7} \times \frac{5}{4}= \frac{3 \times 5}{7 \times 4}=\frac{15}{28}$

You'll be shocked and stunned to learn that Impala does exactly the same thing with algebraic fractions:
$\frac{(x+2)}{(x-4)} \div \frac{x+3}{(2x-3)} = \frac{(x+2)}{(x-4)} \times \frac{(2x-3)}{(x+3)} = \frac{(x+2)(2x-3)}{(x-4)(x+3)}$

Rule #3: Add and subtract - bottoms the same

OK, this is the one you've probably struggled with in the past. But no longer. You know how to do this. The easiest way is to times each fraction (top and bottom) by the bottom of the other - then do the relevant sum with the tops. Let's try it:
$\frac{3}{7}+\frac{4}{5}=\frac{3}{7} \times \frac{5}{5} + \frac{4}{5} \times \frac{7}{7}= \frac{15}{35}+\frac{28}{35}=\frac{43}{35}$

What do you know? The malarkey you have to go through with algebraic fractions... is just the same.
$\frac{(x+2)}{(x-4)}+\frac{x+3}{(2x-3)}=\frac{(x+2)(2x-3)}{(x-4)(2x-3)}+\frac{(x+3)(x-4)}{(2x-3)(x-4)}=\frac{(x+2)(2x-3)+(x+3)(x-4)}{(x-4)(2x-3)}$
(and you can tidy up the top later).

Quick aside about manners

What's all this about tidying up? What makes some sums tidier than others? With algebraic fraction, the general rule is that you want things to be simplified then put in brackets as much as possible. That last example? The bottom is tidy, but the top could use some work. Let's split it off and work it through:
$(x+2)(2x-3)=2x^2+x-6$
$(x+3)(x-4)=x^2-x-12$
$(x+2)(2x-3)+(x+3)(x-4)=3x^2-18=3(x^2-6)$

So the full, tidied-up fraction would be:
$\frac{3(x^2-6)}{(x-4)(2x-3)}$

Rule #4: Brackets, then cancel

Last rule, the one most people try to break and lose embarrassing marks over. It's very easy to avoid this humiliation: don't cancel anything that's not in a bracket. Be paranoid about it. Make $x^2$ into $(x)(x)$. Make 12 into $(3)(2)(2)$, and then cross the brackets out in pairs - one on top, one on the bottom. If I catch you screwing this up, I will mock you.

$\frac{12}{54}=\frac{(2)(2)(3)}{(2)(3)(3)(3)}=\frac{2}{(3)(3)}=\frac{2}{9}$

I'll even mock you over algebra. I'm warning you.

$\frac{x^2+5x+4}{2x^2+x-3}=\frac{(x+4)(x+1)}{(2x-3)(x+1)}=\frac{x+4}{2x-3}$

Easy enough, right? If you can just think back to these four rules whenever you're dealing with a fraction and you'll be fine nine times out of ten. Or $\frac{9}{10}$ of the time, if you must.

(By the way: Impala wants to have a boat when he's rich. That's why there's a boat in the picture.)

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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