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?
+128448 [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
+26367 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 = ?
\(\begin{array}{|lrcll|} \hline (1)+(3)-(2): & \mathbf{\dfrac{3}{5}}+\mathbf{\dfrac{1}{4}}-\mathbf{ \dfrac{2}{3}} &=& \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) + \left( \mathbf{\dfrac{1}{z} + \dfrac{1}{y}} \right) - \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) \\\\ & \dfrac{3}{5} +\dfrac{1}{4} - \dfrac{2}{3} &=& \dfrac{2}{y} \\\\ & \dfrac{2}{y} &=& \dfrac{3}{5} +\dfrac{1}{4} - \dfrac{2}{3} \\\\ & \dfrac{2}{y} &=& \dfrac{ 3\cdot 12+15-2\cdot 20}{60} \\\\ & \dfrac{1}{y} &=& \dfrac{36+15-40}{120} \\\\ & \dfrac{1}{y} &=& \dfrac{11}{120} \\\\ & \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{120}{11}} \\ \hline \end{array}\)
