0
$\begingroup$

Good evening guys, ok so ive been stumped on this for some time and am a bit rusty with my fractions so i was wondering if i could get some help on answering this

I have to try and write this fraction on its own and in its most basic form (simple)

The problem:

$\dfrac{\big(\frac{a^3 - 3a^2}{8a^2 - 4a}\big)}{\big(\frac{a^2 - 9}{2a^2 + 5a - 3}\big)}$

Am really stuck, any help would be appreciated!

$\endgroup$
  • $\begingroup$ They key with these monstrosities is always to factor each piece first, and then pray for cancellation. Try that, and get back to us :P $\endgroup$ – Alfred Yerger Sep 8 '16 at 1:04
  • $\begingroup$ Factor what can be factored. Cancel what can be canceled. Then the big piece. Multiply top and bottom by the denominator of what is on the bottom. This will get rid of the fraction on the bottom. The multiply top and bottom by the denominator of the fraction that is in the numerator. This will give you one fraction. $\endgroup$ – Doug M Sep 8 '16 at 1:05
  • $\begingroup$ The whole thing seems to be a pain, am just unsure about how this is done, but ill follow your advice, thanks! $\endgroup$ – Zochonis Sep 8 '16 at 1:15
0
$\begingroup$

When dividing something by a fraction, we multiply by the reciprocal of the fraction. So we have:

$$\frac{\frac{a^3-3a^2}{8a^2-4a}}{\frac{a^2-9}{2a^2+5a-3}}=\frac{a^3-3a^2}{8a^2-4a} \times \frac{2a^2+5-3}{a^2-9}$$

Here it can help to factor out terms early on:

$$\frac{a^3-3a^2}{8a^2-4a} \times \frac{2a^2+5-3}{a^2-9}=\frac{a^2(a-3)}{4a(2a-1)} \times \frac{(a+3)(2a-1)}{(a+3)(a-3)}$$

We can now eliminate some terms that appear in both the numerator and denominator. Simplifying, we get:

$$\frac{a^2(a-3)}{4a(2a-1)} \times \frac{(a+3)(2a-1)}{(a+3)(a-3)}=\frac{a}{4}$$

$\endgroup$
  • $\begingroup$ What did you eliminate to get straight to $\frac{a}{4}$ ? $\endgroup$ – Zochonis Sep 8 '16 at 1:23
  • $\begingroup$ The $(a-3)$ cancels out, the $(a+3)$ cancels out, the $(2a-1)$ cancels out, and we are left with $\frac{a^2}{4a} = \frac{a}{4}$. $\endgroup$ – user2825632 Sep 8 '16 at 1:25
  • $\begingroup$ Ah i see now, so in any fraction like this (a lot of algebra) and also is a division would this be the best method to use? $\endgroup$ – Zochonis Sep 8 '16 at 1:36
  • $\begingroup$ Yes - I suggest factoring before you do any multiplication or division, since it's easier to cancel out terms. $\endgroup$ – user2825632 Sep 8 '16 at 1:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.