+0

# Math Challenge #something + 1

0
117
3
+302

The last one was quite easy, but for the benefit of the people who didn't get it, here's the solution:

$$a+b=10, a^2+b^2=44,a^3+b^3=?$$

$$(a+b)^2=10^2$$

$$a^2+b^2+2ab=100$$

$$44+2ab=100$$

$$2ab=56,ab=28$$

$$a^3+b^3=(a+b)(a^2-ab+b^2)$$

$$a^3+b^3=(10)(44-28)$$

$$a^3+b^3=10*16$$

$$a^3+b^3=160$$

For those who solved it, good job! Now for the next one:

$$wxy=10$$

$$wyz=5$$

$$wxz=45$$

$$xyz=12$$

Find $$w+x+y+z$$

GOOD LUCK!

Mathhemathh  Aug 15, 2017
Sort:

#1
+12
+1

$$12 {1\over2}$$

DarDragon  Aug 15, 2017
#2
0

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

Guest Aug 15, 2017
#3
+18712
0

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.

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

heureka  Aug 16, 2017
edited by heureka  Aug 16, 2017

### 19 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details