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-2
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avatar+133 

The system of equations \(\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 3\)
has one ordered triple solution (x,y,z). What is the value of in this solution?

 Mar 25, 2020
edited by CPhilFanboy  Mar 25, 2020
 #1
avatar+1956 
+2

Hint:

 

Try making all of these fractions the same denominator~

 Mar 25, 2020
 #2
avatar+133 
+2

How do I do that?

CPhilFanboy  Mar 25, 2020
 #3
avatar+1956 
+2

Multiply all of these by a common denominator.

 

I'm not 100% this is correct, though...

CalTheGreat  Mar 25, 2020
 #4
avatar+133 
+2

There isn't a common denominator.

CPhilFanboy  Mar 25, 2020
 #5
avatar+1956 
+2

There is. You can multiply it by a certain thing.

 

Anyway, IDK if this is correct

CalTheGreat  Mar 25, 2020
 #6
avatar+133 
+2

This is urgent I really need help right now, do you know the answer, or can you teach me how to do it and tell the answwer?

CPhilFanboy  Mar 25, 2020
 #7
avatar+111321 
+2

xy                             xz                          yz

_____    = 1        _______  =   2          _______   =   3

x + y                      x + z                        y + z

 

We can  convert these to this system:

x + y  =  xy      ⇒  xy - y  = x  ⇒   y ( x - 1)  = x   ⇒  y  = x  / (x - 1)         (1) 

2x + 2z = xz    ⇒   xz - 2z =  2x ⇒  z ( x - 2)  = 2x  ⇒  z   =  2x / (x - 2)     (2)

3y + 3z =  yz     (3)

 

Sub (1)  and (2)  into 3

 

3x / (x  -1)  + 3 (2x)  /(x - 2)   = (x * 2x) / [ (x - 1)(x - 2)]

 

Multiply through  by (x  -1) (x - 2)

 

3x ( x -2)  + 6x ( x -1)  =  2x^2        simplify

 

3x^2 - 6x  + 6x^2  - 6x = 2x^2

 

7x^2   - 12x  = 0      factor

 

x ( 7x  - 12)  = 0

 

The first factor   x  = 0  provides  no  solution

 

Setting the  second factor to 0  and solving for x produces  x = 12/7

 

And  y  = (12/7) / (12/7 -1)  =  (12/7) / (5/7)  =  12/5

 

And  z =   2 (12/7)/ (12/7 - 2)  =  (24/7) / (-2/7)  =  24/ -2   = -12

 

 

cool cool cool

 Mar 26, 2020
 #8
avatar+23566 
+1

Strong work, Chris !    ouch ! cheeky

ElectricPavlov  Mar 26, 2020
 #9
avatar+111321 
0

Thanks, EP   !!!!!

 

 

cool cool cool

CPhill  Mar 26, 2020
 #10
avatar+24949 
+1

The system of equations \(\dfrac{xy}{x + y} = 1, \quad \dfrac{xz}{x + z} = 2, \quad \dfrac{yz}{y + z} = 3\)
has one ordered triple solution \((x,y,z)\).

What is the value of in this solution?

 

\(\begin{array}{|rcll|} \hline \dfrac{xy}{x + y} &=& 1 \\\\ \dfrac{x + y}{xy} &=& 1 \\\\ \dfrac{x}{xy}+\dfrac{y}{xy} &=& 1 \\\\ \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} &=& \mathbf{1} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{xz}{x + z} &=& 2 \\\\ \dfrac{x + z}{xz} &=& \dfrac{1}{2} \\\\ \dfrac{x}{xz}+\dfrac{z}{xz} &=& \dfrac{1}{2} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} &=& \mathbf{\dfrac{1}{2}} \qquad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{yz}{y + z} &=& 3 \\\\ \dfrac{y + z}{yz} &=& \dfrac{1}{3} \\\\ \dfrac{y}{yz}+\dfrac{z}{yz} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} &=& \mathbf{\dfrac{1}{3}} \qquad (3) \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline -(1)+(2)+(3) : & -\dfrac{1}{y}-\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{y} &=& -1+\dfrac{1}{2}+\dfrac{1}{3} \\\\ & \dfrac{2}{z} &=& \dfrac{-1}{6} \\\\ & \dfrac{z}{2} &=& -6 \\\\ & \mathbf{z} &=& \mathbf{-12} \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)-(2)+(3) : & \dfrac{1}{y}+\dfrac{1}{x}- \dfrac{1}{z}-\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{y} &=& 1-\dfrac{1}{2}+\dfrac{1}{3} \\\\ & \dfrac{2}{y} &=& \dfrac{5}{6} \\\\ & \dfrac{y}{2} &=& \dfrac{6}{5} \\\\ & \mathbf{y} &=& \mathbf{\dfrac{12}{5}} \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)+(2)-(3) : & \dfrac{1}{y}+\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{x}-\dfrac{1}{z}-\dfrac{1}{y} &=& 1+\dfrac{1}{2}-\dfrac{1}{3} \\\\ & \dfrac{2}{x} &=& \dfrac{7}{6} \\\\ & \dfrac{x}{2} &=& \dfrac{6}{7} \\\\ & \mathbf{x} &=& \mathbf{\dfrac{12}{7}} \\ \hline \end{array} \)

 

laugh

 Mar 26, 2020

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