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# System of equations

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Assuming x, y and z are real positive numbers satisfying:

\begin{align*} xy-z&=15, \\ xz-y&=0, \text{ and} \\ yz-x&=0, \end{align*}

then, what is the value of xyz?

Guest Jul 11, 2018
#1
+20151
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Assuming x, y and z are real positive numbers satisfying:

\begin{align*} xy-z&=15, \\ xz-y&=0, \text{ and} \\ yz-x&=0, \end{align*}
\begin{align*} xy-z&=15, \\ xz-y&=0, \text{ and} \\ yz-x&=0, \end{align*}
then, what is the value of xyz?

$$\begin{array}{|lrcll|} \hline & xy-z&=& 15 \\ (1) & \mathbf{xy} &\mathbf{=}& \mathbf{15 + z} \\\\ & xz-y&=& 0 \\ (2) & \mathbf{xz} & \mathbf{=}& \mathbf{y} \\\\ & yz-x&=0 \\ (3) & \mathbf{yz} & \mathbf{=}& \mathbf{x} \\\\ \hline (2)+(3): & xz + yz &=& y + x \\ & z(x+y) &=& x+y \\ & z &=& \dfrac{x+y}{x+y} \\ & \mathbf{z} & \mathbf{=}& \mathbf{1} \\\\ \hline & xy & = & 15 + z \quad & | \quad z = 1 \\ & xy & = & 15 + 1 \\ & \mathbf{xy} &\mathbf{=}& \mathbf{16} \\\\ & xyz & = & xy\cdot z \quad & | \quad xy = 16 \qquad z = 1 \\ & xyz & = & 16\cdot 1 \\ & \mathbf{xyz} &\mathbf{=}& \mathbf{16} \\ \hline \end{array}$$

heureka  Jul 11, 2018