Tricky Equations

Algebra Level 4

$\begin{aligned}xyz=&3\\x+\dfrac1z=&4\\y+\dfrac1x=&5\\z+\dfrac1y=&\dfrac ab\end{aligned}$

where $a$ and $b$ are coprime positive integers, find the value of $a+b$ .

1 solution

A Former Brilliant Member
Apr 20, 2016

$\rightarrow x+\dfrac{1}{z}=4$ .... $(1)$

$\rightarrow y+\dfrac{1}{x}=5$ .... $(2)$

$\rightarrow z+\dfrac{1}{y}=\dfrac{a}{b}$ .... $(3)$

Adding $(1)$ , $(2)$ and $(3)$ .

$\Rightarrow x+y+z+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}=\color{#3D99F6}{9+\dfrac{a}{b}}$

Multiplying $(1)$ , $(2)$ and $(3)$ .

$\Rightarrow \left(x+\dfrac{1}{z}\right)×\left(y+\dfrac{1}{x}\right)×\left(z+\dfrac{1}{y}\right)=\dfrac{20a}{b}$

$xyz+\dfrac{1}{xyz}+\left(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{20a}{b}$

$3+\dfrac{1}{3}+\left(\color{#3D99F6}{9+\dfrac{a}{b}}\right)=\dfrac{20a}{b}$

$\dfrac{37}{3}+\dfrac{a}{b}=\dfrac{20a}{b}$

$\dfrac{a}{b}=\dfrac{37}{57}$

$\therefore a+b=37+57=\boxed{94}$

x=13/8 , y=57/13 , z=24/57. substituting them into the question gives a/b=37/57

Aditya Kumar - 5 years, 1 month ago