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Eliminate $x,y,z$ between the equations
$$\dfrac{y}{z}-\dfrac{z}{y}=a,\dfrac{z}{x}-\dfrac{x}{z}=b,\dfrac{x}{y}-\dfrac{y}{x}=c$$.

I understand that if I can somehow find the values of $\dfrac{x}{y},\dfrac{y}{z},\dfrac{z}{x}$ in terms of $a,b,c$ then I am done by $\dfrac{x}{y}\cdot\dfrac{y}{z}\cdot\dfrac{z}{x}$ but I cannot seem to do this. I didn't want to multiply the given statements when I approached the problem as I thought that, instead of helping, this would create unnecessary extension and complication of the problem. Please help.

Hawk
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1 Answers1

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Let $r=y/z,\ s=z/x,\ t=x/y.$ Then $r-1/r=a,\ s-1/s=b,\ t-1/t=c.$ The first two have each two solutions for $r,s$ which are $$r=\frac{a \pm \sqrt{a^2+4}}{2},\ \ s=\frac{b \pm \sqrt{b^2+4}}{2}.\tag{1}$$ The only restriction on $r,s,t$ is that $rst=1,$ i.e. each of $r,s$ may be chosen arbitrarily (nonzero), and then $t$ is determined as $t=1/(rs).$ Here the values of $r,s$ are, from a given pair $a,b$, expressed by one of the sign choices in $(1).$

So if $a,b$ are chosen, and also one of the four sign choices in $(1)$ is made, the value of $c$ is determined as $c=1/(rs)-rs.$

coffeemath
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