[SOLVED]GCSE: Simplifying Fractions

i think you mean the "denominator" has a factor of x-2 and the "numerator" has a factor of 2-x.... i think that's been mixed up there.

I caught this too and double checked to make sure I copied it correctly from the textbook, I have. Seems like a silly error missed by the editors.


In Step 3, if you multiply out the numerator wouldn't it come to $-2+x$?

As the above post has suggested, multiply top and bottom by -1 to get the result you want!!!! the book is just taking you the long way round

So if I multiply the top and bottom by (-1), to me it should look like this:

[a]

$\displaystyle \frac{(-1)[(2-x)(2+x)]}{(-1)[(x-2)(x-1)]}$

Which is also the same as:



$\displaystyle \frac{(-1)(2-x)(-1)(2+x)}{(-1)(x-2)(-1)(x-1)}$

Which would give:

[c]

$\displaystyle \frac{(-2+x)(-2-x)}{(-x+2)(-x+1)}$

Right?

But doing this just gives me the inverse of the original and I still don't see how they cancel out. I can see where if only $(2-x)$ was multiplied by (-1) it would equal $x+2$ and then it could be cancelled out with the dominator. But I thought the whole of the numerator and denominator has to be multiplied or divided by whatever number you introduce to keep the ratio between the numerator and denominator the same - as in [a] and .

This is my understanding, and I can't figure out the difference between this and what's in the textbook.

Stating:



Isn't really helpful when that's one of the things I'm confused about.

I got it now, I did some 'specializing' and used integers in place of letters and - duh - $2-x$ is the same as $(-1)(x+2)$, it's just another way of writing it - and doing it this way reveals the common factors so that they can be removed, just as you would factorize quadratics in a fraction to determine common factors for simplifying.

2-x = (-1)(x-2) =/= (-1)(x+2)

Woops, your right. Thanks. Fortunately I understand that, I made a typo.

Here's another, which I seem to have got stuck on:

$\displaystyle \frac{a-1}{1-a^2}$.

Here's my working:

[1]

Now $(1-a^2)=(-1)(a^2-1)$, so:

$\displaystyle \frac{a-1}{(-1)(a^2-1)}$ $=$ $\displaystyle \frac{a-1}{(-1)(a \times a-1)}$, which gives:

[2]

$\displaystyle \frac{1}{(-1)(a)}$, which gives:

[3]

$\displaystyle \frac{1}{-a}$

But the answer in the textbook shows:

$\displaystyle \frac{-1}{1+a}$

Where have I made a mistake?

$(a^2-1) = (a+1)(a-1)$

Not a(a-1) as you put. That's the mistake.

So, working it through:

$(1-a^2)=(-1)(a^2-1)$

$\displaystyle \frac{a-1}{(-1)(a^2-1)}$ $=$ $\displaystyle \frac{a-1}{(-1)(a+1)(a-1)}$

$\displaystyle \frac{1}{(-1)(a+1)}$

$\displaystyle \frac{-1}{a+1}$

As in the textbook.