# What am I doing wrong with this simple problem?

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?

• $1/(X/Z + Y/Z)$ is not $(Z/X +Z/Y)$ – Stefan Hansen Jul 1 '14 at 8:43
• ${W\over{X\over Z}+{Y\over Z}}\times{Z\over Z}= {W\times Z\over{X\over Z}\times Z+{Y\over Z}\times Z} = {WZ\over X+Y}$ – Graham Kemp Jul 1 '14 at 8:47
• 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 $\displaystyle{W}\cdot(\frac{Z}{X}+\frac{Z}{Y})=1\cdot(\frac{2}{1}+\frac{2}{1})=4$. – barak manos Jul 1 '14 at 8:52
• 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. – littleO Jul 1 '14 at 11:08

$$A = W \cdot \dfrac{1}{\dfrac{X}{Z}+\dfrac{Y}{Z}}$$
$$\dfrac{1}{\dfrac{X}{Z}+\dfrac{Y}{Z}} \ne \dfrac{Z}{X} + \dfrac{Z}{Y}$$
To see this let $X = 1$, $Y = 2$ and $Z = 3$
$$\dfrac{1}{\dfrac{X}{Z}+\dfrac{Y}{Z}} = \dfrac{1}{\dfrac{1}{3}+\dfrac{2}{3}} = 1 \ne \dfrac{Z}{X} + \dfrac{Z}{Y} = \dfrac{3}{1} + \dfrac{3}{2} = 4.5$$