Question

If x/3x-y-z=y/3y-z-x=z/3z-x-y and x+y+z≠0 then show that the value of each ratio is equal to 1.

Answer 1

Answer:

$\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=1$

Step-by-step explanation:

If x/3x-y-z=y/3y-z-x=z/3z-x-y and x+y+z≠0 then show that the value of each ratio is equal to 1

Concept used:

$\boxed{\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\:\implies\:\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}}$

$\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}$

$\implies\:\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{3x-y-z+3y-z-x+3z-x-y}$

$\implies\:\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{3x+3y+3z-2x-2y-2z}$

$\implies\:\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{3(x+y+z)-2(x+y+z)}$

$\implies\:\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{x+y+z}$

$\implies\:\boxed{\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=1}$

Hence proved

Answer 2

Answer:

this is your answer.

Hope it will help you.