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     (a) make x the subject of the formular  $${\frac{{\mathtt{1}}}{{\mathtt{z}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{x}}}{{\mathtt{y}}}} = {\frac{{\mathtt{5}}}{{\mathtt{y}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y}}\right)}}$$ ,

      (b) next,if x:y:z=2:5:3 , find the value(s) of x satisfyinh the formular in (a)

 

I found the answer in (a) is $${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}{{\mathtt{\,-\,}}\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}}$$ , how to solve (b)?

 Jan 14, 2015

Best Answer 

 #3
avatar+118724 
+8

 next,if x:y:z=2:5:3 , find the value(s) of x satisfyinh the formular in (a)

 

$$\\y=\frac{x}{2}*5=\frac{5x}{2}, \qquad z=\frac{x}{2}*3=\frac{3x}{2}\\\\\\
x=\frac{10z-2y}{5z}\\\\
x=\frac{10*\frac{3x}{2}-2*\frac{5x}{2}}{5*\frac{3x}{2}}\\\\
x=\frac{15x-5x}{\frac{15x}{2}}\\\\
x=10x*\frac{2}{15x}\qquad \mbox{Note x cannot equal 0}\\\\
x=\frac{4}{3}$$

 Jan 14, 2015
 #1
avatar+118724 
+5

I think that your part a question can be simplified further to

 

$$x=\dfrac{10z-2y}{5z}$$

 Jan 14, 2015
 #2
avatar+26400 
+5

I found the answer in (a) is $${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}{{\mathtt{\,-\,}}\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}}$$ , how to solve (b)?

$$x=\dfrac{10zy-2y^2}{5zy} = 2-\dfrac{2}{5}*\dfrac{y}{z} \quad | \quad \dfrac{y}{z} = \dfrac{5}{3}\\\\
x= 2 - \frac{2}{3} = 1\frac{1}{3}$$

 Jan 14, 2015
 #3
avatar+118724 
+8
Best Answer

 next,if x:y:z=2:5:3 , find the value(s) of x satisfyinh the formular in (a)

 

$$\\y=\frac{x}{2}*5=\frac{5x}{2}, \qquad z=\frac{x}{2}*3=\frac{3x}{2}\\\\\\
x=\frac{10z-2y}{5z}\\\\
x=\frac{10*\frac{3x}{2}-2*\frac{5x}{2}}{5*\frac{3x}{2}}\\\\
x=\frac{15x-5x}{\frac{15x}{2}}\\\\
x=10x*\frac{2}{15x}\qquad \mbox{Note x cannot equal 0}\\\\
x=\frac{4}{3}$$

Melody Jan 14, 2015
 #4
avatar+118724 
0

You beat me by a hair heureka     

 Jan 14, 2015
 #5
avatar+26400 
+5

Hi,  Melody

excuse please

 Jan 14, 2015
 #6
avatar+130536 
+5

1/z + x/y = 5/y - 3x / 2y     rearrange

3x / 2y  + x / y  = 5/y - 1/z    simplify

5x / 2y  = [5z - y ] / yz

5x = 2[5z - y ] / z

x  = (2/5) [ 5z - y ] / z = [ 10z - 2y ] /  [5z] = 2 - 2y/5z = 2 -  (2/5) (y/z)....but (y/z)  = 5/3..so we have

x =2 - (2/5)(5/3)  =  2 - 2/3  = 6/3 - 2/3 = 4/3

 

 Jan 14, 2015
 #7
avatar+118724 
+3

There is nothing to excuse Heureka, you beat me fair and square.  LOL  

 Jan 14, 2015

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