Given $\frac {a\cdot y}{b\cdot x} = \frac CD$, find $y$.

That's a pretty easy one... I have the following equality : $\dfrac {a\cdot y}{b\cdot x} = \dfrac CD$ and I want to leave $y$ alone so I move "$b\cdot x$" to the other side

$$a\cdot y= \dfrac {(C\cdot b\cdot x)}{D}$$

and then "$a$"

$$y=\dfrac {\dfrac{(C\cdot b\cdot x)}D} a.$$

Where is my mistake? I should be getting $y= \dfrac {(b\cdot C\cdot x)}{(a\cdot D)}$.

I know that the mistake I am making is something very stupid, but can't work it out. Any help? Cheers!

• (+1) Thanks for showing your work. It makes it much easier to help someone this way. Everyone really ought to up vote these questions more. – user70962 Jun 9 '13 at 8:06

No mistake was made. Observe that: $$y=\dfrac{\left(\dfrac{Cbx}{D}\right)}{a}=\dfrac{Cbx}{D} \div a = \dfrac{Cbx}{D} \times \dfrac{1}{a}=\dfrac{Cbx}{Da}=\dfrac{bCx}{aD}$$ as desired.
Note that $$\frac{\dfrac{1}{x}}{y} = \frac{1}{x}\frac{1}{y} = \frac{1}{xy}.$$ This process is often called 'invert and multiply'. If you apply this rule to your final expression, you will find it agrees with the given solution.
More generally $$\frac{\dfrac{a}{b}}{\dfrac{c}{d}} = \frac{a}{b}\frac{d}{c} = \frac{ad}{bc}.$$