1. ## Halfway through equation

I need h in terms of x

Got up to

$\frac{4}{3} \cdot 2x=\frac{1}{3} \cdot h$

2. Multiply both sides with 3 to obtain: $8x = h$

That is what you want right?

3. Oh god. I came up with $\frac{\frac{4}{3} 2x}{\frac{1}{3}}$

Thanks though, I'm so forgetful ^^

4. Oh god. I came up with
That is correct: $\frac{\frac{4}{3}2x}{\frac{1}{3}} = \frac{4\times \frac{1}{3}\times 2x}{\frac{1}{3}}= 8x$

5. yes but it's probably not what the examiner wanted! XD

6. Dividing by $\frac{1}{3}$ is multiplying by $3.$

For example....

Try dividing 12 into 3 equal parts, $\frac{12}{3}=4$

Now try dividing 12 into $\frac{1}{3}$ parts...

Make sense?
thought not...

What is 12 divided by $\frac{1}{3}$ ?

$\frac{12}{\frac{1}{3}}=\frac{(36)\frac{1}{3}}{\fra c{1}{3}}=36\frac{\frac{1}{3}}{\frac{1}{3}}$

Expressing the numerator as a multiple of the denominator allows cancellation
of non-zero common factors (after all, a third is 0.67ish).

Any number divided by itself is 1, except zero.
A fraction divided by itself is 1
So, 12 divided by a third is 36, because it's really 12 times 3.

Dividing by a fraction is multiplying by the fraction turned upside down.

of course, this means... multiply by the numerator and divide by the denominator.

7. Amazing. I really do love your ways of explaining things. Please do continue to help me in further issues if you find time

8. Thanks, Mukilab,

you're welcome.

The vectors question you had a few days ago......

The critical thing to understand there was $\overrightarrow{xy}=\overrightarrow{y}-\overrightarrow{x}$

The vector $\overrightarrow{xy}$ starts at the point x and ends at y.

The vectors $\overrightarrow{x}$ and $\overrightarrow{y}$ both start at the origin.

You understand this by examining the parallelograms.