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The system of equations $\frac{3xy}{x + y} = 5, \quad \frac{2xz}{x + z} = 3, \quad \frac{yz}{y + z} = 4$ has one ordered triple solution $(x,y,z)$. What is the value of $z$ in this solution?

Guest Oct 29, 2017
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#1
+78590
+2

[3xy] / [x + y ]  = 5

[2xz] / [ x + z ]  = 3

[yz ] / [ y + z ]  = 4  ⇒   yz  =  4 [ y + z ]   (1)

3xy  =  5x + 5y   ⇒  3xy - 5y  = 5x  ⇒    y  [ 3x - 5]  = 5x  ⇒  y = [5x] / [ 3x - 5]     (2)

2xz  =  3x + 3z  ⇒   2xz - 3z  = 3x  ⇒    z [ 2x - 3]  = 3x  ⇒   z = [ 3x] / [ 2x - 3]    (3)

Sub  (2)  and (3)  into (1)

[5x] / [ 3x - 5]  *   [ 3x] / [ 2x - 3]  =   4 [  [5x] / [ 3x - 5]   +  [ 3x] / [ 2x - 3]   ]   simplify

15x^2   =     4 [  5x[2x - 3] + 3x [ 3x - 5 ] ]

15x^2  =  4  [ 10x^2 - 15x  +  9x^2 - 15x ]

15x^2  = 76x^2  - 120x

61x^2   -  120x  = 0

x [ 61x - 120] =  0

So  x = 0    [reject as it makes an original denominator  = 0 ]

Or

61x - 120  = 0   ⇒     x = 120 / 61

So......   z = [ 3x] / [ 2x - 3]

z =   3 (120 / 61]  /  [  2(120/61)  - 3 ]

z =  [ 360 / 61]  / [ 240/61  - 183/61]

z =  [ 360]  / [ 240 - 183 ]

z =  360 / 57  =     120 / 19

CPhill  Oct 29, 2017
#2
+18712
+1

The system of equations $\frac{3xy}{x + y} = 5, \quad \frac{2xz}{x + z} = 3, \quad \frac{yz}{y + z} = 4$

has one ordered triple solution $(x,y,z)$.

What is the value of $z$ in this solution?

$$\begin{array}{|rcll|} \hline \dfrac{3xy}{x + y} = 5, \quad \dfrac{2xz}{x + z} = 3, \quad \dfrac{yz}{y + z} = 4 \\ z=\ ? \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \dfrac{3xy}{x + y} &=& 5 \\\\ & \dfrac{3}{5} &=& \dfrac{x+y}{xy} \\\\ \mathbf{(1)} & \mathbf{\dfrac{3}{5}} &\mathbf{=}& \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \\\\ \hline & \dfrac{2xz}{x + z} &=& 3 \\\\ & \dfrac{2}{3} &=& \dfrac{x+z}{xz} \\\\ \mathbf{(2)} & \mathbf{ \dfrac{2}{3}} &\mathbf{=}& \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \\\\ \hline & \dfrac{yz}{y + z} &=& 4 \\ & \dfrac{1}{4} &=& \dfrac{y+z}{yz} \\\\ \mathbf{(3)} & \mathbf{\dfrac{1}{4}} &\mathbf{=}& \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} \\ \hline \end{array}$$

z = ?

$$\begin{array}{|lrcll|} \hline (2)+(3)-(1): & \mathbf{ \dfrac{2}{3}}+\mathbf{\dfrac{1}{4}} - \mathbf{\dfrac{3}{5}} &=& \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) + \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} \right) - \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) \\\\ & \dfrac{2}{3} +\dfrac{1}{4} - \dfrac{3}{5} &=& \dfrac{2}{z} \\\\ & \dfrac{2}{z} &=& \dfrac{2}{3} +\dfrac{1}{4} - \dfrac{3}{5} \\\\ & \dfrac{2}{z} &=& \dfrac{2\cdot 20+15-3\cdot 12}{60} \\\\ & \dfrac{1}{z} &=& \dfrac{40+15-36}{120} \\\\ & \dfrac{1}{z} &=& \dfrac{19}{120} \\\\ & \mathbf{z} &\mathbf{=}& \mathbf{\dfrac{120}{19}} \\ \hline \end{array}$$

x = ?

$$\begin{array}{|lrcll|} \hline (1)-(3)+(2): & \mathbf{\dfrac{3}{5}}-\mathbf{\dfrac{1}{4}}+\mathbf{ \dfrac{2}{3}} &=& \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) - \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} \right) + \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) \\\\ & \dfrac{3}{5} -\dfrac{1}{4} + \dfrac{2}{3} &=& \dfrac{2}{x} \\\\ & \dfrac{2}{x} &=& \dfrac{3}{5} -\dfrac{1}{4} + \dfrac{2}{3} \\\\ & \dfrac{2}{x} &=& \dfrac{ 3\cdot 12-15+2\cdot 20}{60} \\\\ & \dfrac{1}{x} &=& \dfrac{36-15+40}{120} \\\\ & \dfrac{1}{x} &=& \dfrac{61}{120} \\\\ & \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{120}{61}} \\ \hline \end{array}$$

y = ?

$$\begin{array}{|lrcll|} \hline (1)+(3)-(2): & \mathbf{\dfrac{3}{5}}+\mathbf{\dfrac{1}{4}}-\mathbf{ \dfrac{2}{3}} &=& \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) + \left( \mathbf{\dfrac{1}{z} + \dfrac{1}{y}} \right) - \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) \\\\ & \dfrac{3}{5} +\dfrac{1}{4} - \dfrac{2}{3} &=& \dfrac{2}{y} \\\\ & \dfrac{2}{y} &=& \dfrac{3}{5} +\dfrac{1}{4} - \dfrac{2}{3} \\\\ & \dfrac{2}{y} &=& \dfrac{ 3\cdot 12+15-2\cdot 20}{60} \\\\ & \dfrac{1}{y} &=& \dfrac{36+15-40}{120} \\\\ & \dfrac{1}{y} &=& \dfrac{11}{120} \\\\ & \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{120}{11}} \\ \hline \end{array}$$

heureka  Oct 30, 2017

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