Question

х
(ii) If
Z
у
and x+y +z #0 then show that the value of
3x - y - z 3y - z - x
3y - z-x 3z - x - y
each ratio is equal to 1.​

Answer 1

ANSWER:

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}}xa=yb=zc⟹xa=yb=zc=x+y+za+b+c

\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}3x−y−zx=3y−z−xy=3z−x−yz

\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}⟹3x−y−zx=3y−z−xy=3z−x−yz=3x−y−z+3y−z−x+3z−x−yx+y+z

\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}⟹3x−y−zx=3y−z−xy=3z−x−yz=3x+3y+3z−2x−2y−2zx+y+z

\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)}⟹3x−y−zx=3y−z−xy=3z−x−yz=3(x+y+z)−2(x+y+z)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}⟹3x−y−zx=3y−z−xy=3z−x−yz=x+y+zx+y+z

\implies\:\boxed{\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=1}⟹3x−y−zx=3y−z−xy=3z−x−yz=1

Hence proved

Step-by-step explanation:

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