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# system

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

xy - z = 15

xz - y = 15

yz - x = 15

then, what is the value of  xyz?

$$\phantom{yz - x = 0}$$

$$\phantom{xz - y = 0}$$

Jul 6, 2022

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Hi Guest!

Rewrite the first two equations as follows:

$$xy=15+z$$  (1)

$$xz=15+y$$  (2)

Subtracting (2) from (1) gives:

$$x(y-z)=-(y-z)$$       (Notice: $$z-y=-(y-z)$$ ).

$$x(y-z)+(y-z)=0$$        (Add (y-z) to both sides to make R.H.S = 0 , and factor (y-z)):

$$(y-z)(x+1)=0$$

Thus, $$y=z, \text{or } x=-1 (rejected)$$  (As given x,y,z > 0).

Therefore, $$y=z$$ only.

Also, using the second and third equation:

$$xz=15+y$$  (2)

$$yz=15+x$$  (3)

Subtracting (3) from (2) to get:

$$z(x-y)=-(x-y)$$

$$z(x-y)+(x-y)=0$$

$$(x-y)(z+1)=0$$

Therefore, $$x=y$$        (z>0 so z=-1 is rejected).

But we found that:  $$y=z$$

Thus, It must mean that they are all equal to each other. That is,  $$x=y=z$$

So our system of equations becomes:

$$x^2-x=15$$ $$\iff x^2-x-15=0 \iff x=\frac{1+\sqrt{61}}{2}$$  (A quadratic equation in x, which can be solved by the quadratic formula.)

$$xyz=x^3=\frac{(1+\sqrt{61})^3}{8}=23+8\sqrt{61}$$