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# Math Challenge #something + 1

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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
+18564
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

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