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?

  • 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

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 $Z = 3$

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


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