X
Z
If
y
3x-y-z 3y-z-x 32-x- y
each ratio is equal to 1.
and x+y +z = 0 then show that the value of
Answers
Answer:
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}}
x
a
=
y
b
=
z
c
⟹
x
a
=
y
b
=
z
c
=
x+y+z
a+b+c
\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}
3x−y−z
x
=
3y−z−x
y
=
3z−x−y
z
\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−z
x
=
3y−z−x
y
=
3z−x−y
z
=
3x−y−z+3y−z−x+3z−x−y
x+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−z
x
=
3y−z−x
y
=
3z−x−y
z
=
3x+3y+3z−2x−2y−2z
x+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−z
x
=
3y−z−x
y
=
3z−x−y
z
=
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−z
x
=
3y−z−x
y
=
3z−x−y
z
=
x+y+z
x+y+z
\implies\:\boxed{\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=1}⟹
3x−y−z
x
=
3y−z−x
y
=
3z−x−y
z
=1