+223 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?
+2094 Multiply all of these by a common denominator.
I'm not 100% this is correct, though...
+2094 There is. You can multiply it by a certain thing.
Anyway, IDK if this is correct
+223 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?
+128408 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
+26367 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?
\(\begin{array}{|rcll|} \hline \dfrac{xy}{x + y} &=& 1 \\\\ \dfrac{x + y}{xy} &=& 1 \\\\ \dfrac{x}{xy}+\dfrac{y}{xy} &=& 1 \\\\ \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} &=& \mathbf{1} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{xz}{x + z} &=& 2 \\\\ \dfrac{x + z}{xz} &=& \dfrac{1}{2} \\\\ \dfrac{x}{xz}+\dfrac{z}{xz} &=& \dfrac{1}{2} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} &=& \mathbf{\dfrac{1}{2}} \qquad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{yz}{y + z} &=& 3 \\\\ \dfrac{y + z}{yz} &=& \dfrac{1}{3} \\\\ \dfrac{y}{yz}+\dfrac{z}{yz} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} &=& \mathbf{\dfrac{1}{3}} \qquad (3) \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline -(1)+(2)+(3) : & -\dfrac{1}{y}-\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{y} &=& -1+\dfrac{1}{2}+\dfrac{1}{3} \\\\ & \dfrac{2}{z} &=& \dfrac{-1}{6} \\\\ & \dfrac{z}{2} &=& -6 \\\\ & \mathbf{z} &=& \mathbf{-12} \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)-(2)+(3) : & \dfrac{1}{y}+\dfrac{1}{x}- \dfrac{1}{z}-\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{y} &=& 1-\dfrac{1}{2}+\dfrac{1}{3} \\\\ & \dfrac{2}{y} &=& \dfrac{5}{6} \\\\ & \dfrac{y}{2} &=& \dfrac{6}{5} \\\\ & \mathbf{y} &=& \mathbf{\dfrac{12}{5}} \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)+(2)-(3) : & \dfrac{1}{y}+\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{x}-\dfrac{1}{z}-\dfrac{1}{y} &=& 1+\dfrac{1}{2}-\dfrac{1}{3} \\\\ & \dfrac{2}{x} &=& \dfrac{7}{6} \\\\ & \dfrac{x}{2} &=& \dfrac{6}{7} \\\\ & \mathbf{x} &=& \mathbf{\dfrac{12}{7}} \\ \hline \end{array} \)
