C2 questions

2. Simplify these expressions:

b) x^-2 ÷ x^-3

d) (2x² ÷ 4y³ )^-2



On 2b I get
$x^-1 =$ $\frac{1}{x}$ the answer is$x$

n 2d I don't understand any of it.. what am I missing out on? Answer is
4y⁶ / x⁴

Note the rule that $a^{-n} = \frac{1}{a^n}$.

So here you're doing $\displaystyle \frac{x^{-2}}{x^{-3}} = \frac{\frac{1}{x^2}}{\frac{1}{x^3}} = \frac{x^3}{x^2}$



Again, same thing here:

$\displaystyle \left(\frac{2x^2}{4y^3}\right)^{-2} = \frac{1}{\left(\frac{2x^2}{4y^3}\right)^{2}} = \left(\frac{4y^3}{2x^2}\right)^2$


Can you take it from here knowing that $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ and $(ab)^n = a^nb^n$?

How did you get
$\displaystyle \frac{x^{-2}}{x^{-3}} = \frac{\frac{1}{x^2}}{\frac{1}{x^3}} =[b] [/b]\frac{x^3}{x^2}$ ? - this part

Standard GCSE fractions, $\displaystyle \dfrac{\dfrac{1}{x^2}}{\dfrac{1}{x^3}} \times \dfrac{x^3}{x^3} = \dfrac{\dfrac{x^3}{x^2}}{\dfrac{x^3}{x^3}} = \dfrac{\dfrac{x^3}{x^2}}{1} = \dfrac{x^3}{x^2}$

Where on Earth did that x^3 come from?

2. Simplify these expressions:

b) x^-2 ÷ x^-3

d) (2x² ÷ 4y³ )^-2



On 2b I get
$x^{-1} =$ $\frac{1}{x}$ the answer is$x$

On 2d I don't understand any of it.. what am I missing out on? Answer is
4y⁶ / x⁴

Oh c'mon, you must remember that when you do things like adding fractions, you can multiply a fraction by 1.

So for example, to do $\displaystyle \frac{1}{2} + \frac{1}{4}$ you'd do $\displaystyle \frac{1}{2} \times \frac{2}{2} = \frac{2}{2\times 2} = \frac{2}{4}$.

This is the same thing here, we're doing $\displaystyle \frac{x^{-2}}{x^{-3}} \times \frac{x^3}{x^3}$.

If you don't like thinking of it like that, then what about $\frac{x^{-2}}{x^{-3}} = \frac{1}{x^2} \times \frac{1}{x^{-3}} = \frac{1}{x^2} \times (x^{-3})^{-1} = \frac{x^{3}}{x^2}$.

Any help?

yes, thanks

Sorry about the quality of the picture! :P Also, where I've added the indices - that was a mistake. It was meant to be multiplying

perfection, simple enough for a baby to understand.. good.

\left(\frac{2x^2}{4y^3}\right)^{-2}=\frac{\left(2x^2\right)^{-2}}{\left(4y^3\right)^{-2}}=\frac{\left(4y^3\right)^{2}}{\left(2x^2\right)^{2}}=\frac{ \left( 2y^3 \right)^{2}}{\left(x^2\right)^{2}}=\frac{4y^6}{x^4}

Fixed: $\left(\frac{2x^2}{4y^3}\right)^{-2}=\frac{\left(2x^2\right)^{-2}}{\left(4y^3\right)^{-2}}=\frac{\left(4y^3\right)^{2}}{\left(2x^2\right)^{2}}=\frac{ \left( 2y^3 \right)^{2}}{\left(x^2\right)^{2}}=\frac{4y^6}{x^4}$