Math Challenge #something + 1

$12 {1\over2}$

Divide and substitute and you should get the following:

w = 2 1/2,  x = 6,  y = 2/3,  z = 3, so have:

w+x+y+z = 2 1/2 + 6 + 2/3 + 3 =12 1/6

Now for the next one:

$\begin{array}{rcr} wxy&=&10 \\ wyz&=&5 \\ wxz&=&45 \\ xyz&=&12 \\ \end{array}$
Find $w+x+y+z$

1.

$\begin{array}{|rcll|} \hline xyz &=& 12 \\ xyz \cdot \frac{wyz}{wxy} \cdot \frac{wxz}{wxy} &=& 12 \cdot \frac{5}{10} \cdot \frac{45}{10} \\ xyz \cdot \frac{z}{x} \cdot \frac{z}{y} &=& 12 \cdot \frac{5}{10} \cdot \frac{45}{10} \\ z^3 &=& 12 \cdot \frac{1}{2} \cdot \frac{9}{2} \\ z^3 &=& 3\cdot 9 \\ z^3 &=& 3^3 \\ \mathbf{z} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}$

2.

$\begin{array}{|rcll|} \hline && w+x+y+z \\ &=& \left( \frac{wxy}{xyz} + \frac{wxy}{wyz} + \frac{wxy}{wxz} \right) \cdot z + z \\ &=& \left( \frac{10}{12} + \frac{10}{5} + \frac{10}{45} \right)\cdot z + z \\ &=& \left( \frac{5}{6} + 2 + \frac{2}{9} \right)\cdot z + z \quad & | \quad z= 3 \\ &=& \left( \frac{5}{6} + 2 + \frac{2}{9} \right)\cdot 3 + 3 \\ &=& \frac{5}{2} + 6 + \frac{2}{3} + 3 \\ &=& 9 + \frac{5}{2} + \frac{2}{3} \\ &=& 9 + \frac{15+4}{6} \\ &=& 9 + \frac{19}{6} \\ &=& 12 + \frac{1}{6} \\ \hline \end{array}$