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