Halfway through equation

• January 7th 2010, 09:34 AM
Mukilab
Halfway through equation
I need h in terms of x

Got up to

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

• January 7th 2010, 10:08 AM
Dinkydoe
Multiply both sides with 3 to obtain: $8x = h$

That is what you want right?
• January 7th 2010, 10:34 AM
Mukilab
Oh god. I came up with $\frac{\frac{4}{3} 2x}{\frac{1}{3}}$

Thanks though, I'm so forgetful ^^
• January 7th 2010, 10:38 AM
Dinkydoe
That is correct: $\frac{\frac{4}{3}2x}{\frac{1}{3}} = \frac{4\times \frac{1}{3}\times 2x}{\frac{1}{3}}= 8x$
• January 7th 2010, 12:19 PM
Mukilab
yes but it's probably not what the examiner wanted! XD
• January 7th 2010, 12:41 PM
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.
• January 7th 2010, 12:45 PM
Mukilab
Amazing. I really do love your ways of explaining things. Please do continue to help me in further issues if you find time :)
• January 7th 2010, 02:05 PM
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.