0
$\begingroup$

When solving for A you can do the simple way of multiplying Z and dividing X+Y.

$$A*{X+Y\over Z}=W$$ $$A={WZ\over X+Y}$$

Or the way no one would ever do... $$A*{X+Y\over Z}=W$$ $$A*\left({X\over Z}+ {Y\over Z}\right)=W$$ $$A ={W\over \left({X\over Z}+ {Y\over Z}\right)}$$ This appears to me to be equal to. $$A ={W* \left({Z\over X}+ {Z\over Y}\right)}$$

Clearly this can not be true. Could someone point me to the rule that I am simply forgetting?

$\endgroup$
  • 5
    $\begingroup$ $1/(X/Z + Y/Z)$ is not $(Z/X +Z/Y)$ $\endgroup$ – Stefan Hansen Jul 1 '14 at 8:43
  • $\begingroup$ ${W\over{X\over Z}+{Y\over Z}}\times{Z\over Z}= {W\times Z\over{X\over Z}\times Z+{Y\over Z}\times Z} = {WZ\over X+Y}$ $\endgroup$ – Graham Kemp Jul 1 '14 at 8:47
  • $\begingroup$ The last line is wrong. Take $W=1,X=1,Y=1,Z=2$ for example: $\displaystyle\frac{W}{\frac{X}{Z}+\frac{Y}{Z}}=\frac{1}{\frac{1}{2}+\frac{1}{2}}=1$, but $\displaystyle{W}\cdot(\frac{Z}{X}+\frac{Z}{Y})=1\cdot(\frac{2}{1}+\frac{2}{1})=4$. $\endgroup$ – barak manos Jul 1 '14 at 8:52
  • $\begingroup$ In words, the reciprocal of the sum is not equal to the sum of the reciprocals. A common error is to mistakenly assume that the blip of the blop is equal to the blop of the blip. $\endgroup$ – littleO Jul 1 '14 at 11:08
0
$\begingroup$

Your error here is that while it's true that

$$A = W \cdot \dfrac{1}{\dfrac{X}{Z}+\dfrac{Y}{Z}}$$

Your next assumption is false as pointed out by Stefan Hansen in comments.

$$\dfrac{1}{\dfrac{X}{Z}+\dfrac{Y}{Z}} \ne \dfrac{Z}{X} + \dfrac{Z}{Y}$$

To see this let $X = 1$, $Y = 2$ and $Z = 3$

$$\dfrac{1}{\dfrac{X}{Z}+\dfrac{Y}{Z}} = \dfrac{1}{\dfrac{1}{3}+\dfrac{2}{3}} = 1 \ne \dfrac{Z}{X} + \dfrac{Z}{Y} = \dfrac{3}{1} + \dfrac{3}{2} = 4.5$$

$\endgroup$

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.