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Question-161214




Question Number 161214 by kapoorshah last updated on 14/Dec/21
Commented by mr W last updated on 14/Dec/21
you can not uniquely determine  three unknowns if only two equations  are given. i.e. xyz is not unique!
$${you}\:{can}\:{not}\:{uniquely}\:{determine} \\ $$$${three}\:{unknowns}\:{if}\:{only}\:{two}\:{equations} \\ $$$${are}\:{given}.\:{i}.{e}.\:{xyz}\:{is}\:{not}\:{unique}! \\ $$
Commented by mr W last updated on 14/Dec/21
(1/x)+(1/y)=a−(1/z)  ((x+y)/(xy))=a−(1/z)  ((a−z)/(xy))=a−(1/z)  xy=((a−z)/(a−(1/z)))  xyz=(((a−z)z)/(a−(1/z)))=(((a−z)z^2 )/(az−1))  you see xyz can get any value you want  by selecting a certain value for z.
$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}={a}−\frac{\mathrm{1}}{{z}} \\ $$$$\frac{{x}+{y}}{{xy}}={a}−\frac{\mathrm{1}}{{z}} \\ $$$$\frac{{a}−{z}}{{xy}}={a}−\frac{\mathrm{1}}{{z}} \\ $$$${xy}=\frac{{a}−{z}}{{a}−\frac{\mathrm{1}}{{z}}} \\ $$$${xyz}=\frac{\left({a}−{z}\right){z}}{{a}−\frac{\mathrm{1}}{{z}}}=\frac{\left({a}−{z}\right){z}^{\mathrm{2}} }{{az}−\mathrm{1}} \\ $$$${you}\:{see}\:{xyz}\:{can}\:{get}\:{any}\:{value}\:{you}\:{want} \\ $$$${by}\:{selecting}\:{a}\:{certain}\:{value}\:{for}\:{z}. \\ $$
Commented by kapoorshah last updated on 14/Dec/21
the answer in “a” term
$${the}\:{answer}\:{in}\:“{a}''\:{term} \\ $$
Commented by mr W last updated on 14/Dec/21
you can′t get xyz uniquely in terms of a.
$${you}\:{can}'{t}\:{get}\:{xyz}\:{uniquely}\:{in}\:{terms}\:{of}\:{a}. \\ $$
Commented by kapoorshah last updated on 14/Dec/21
x, y, z are integer
$${x},\:{y},\:{z}\:{are}\:{integer} \\ $$
Commented by mr W last updated on 14/Dec/21
please complete your question at first!  are x,y,z positive or not?  what about a? is a≥3 or not?
$${please}\:{complete}\:{your}\:{question}\:{at}\:{first}! \\ $$$${are}\:{x},{y},{z}\:{positive}\:{or}\:{not}? \\ $$$${what}\:{about}\:{a}?\:{is}\:{a}\geqslant\mathrm{3}\:{or}\:{not}? \\ $$

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