2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ (+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. $\endgroup$ – user70962 Jun 9 '13 at 8:06
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I see , thanks a lot :) $\endgroup$ – Maximilian1988 Jun 9 '13 at 6:48
4
$\begingroup$

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 $$\frac{\dfrac{a}{b}}{\dfrac{c}{d}} = \frac{a}{b}\frac{d}{c} = \frac{ad}{bc}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.