Arraryeldergox2

2022-06-24

$x+2y+z=5\left(x+y\right)\left(y+z\right)\phantom{\rule{0ex}{0ex}}x+y+2z=7\left(y+z\right)\left(z+x\right)\phantom{\rule{0ex}{0ex}}2x+y+z=6\left(z+x\right)\left(x+y\right).$
Find the value of ${24}^{3}xyz$.

Kaydence Washington

Expert

Assume $x+y=a,y+z=b,z+x=c$ and express both LHS and RHS of all $3$ equations in terms of $a,b,c$. After some simplification, you should be able to get a system of linear equations in $\frac{1}{a},\frac{1}{b},\frac{1}{c}$.
Can you solve it now?

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