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Answer 1

Question:

If   $\dfrac{x+y}{ax+by} =\dfrac{y+z}{ay+bz} =\dfrac{z+x}{az+bx}$, prove that each of the ratio is equal to $\dfrac{2}{a+b}$, unless x + y + z = 0

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

$\dfrac{x+y}{ax+by} =\dfrac{y+z}{ay+bz} =\dfrac{z+x}{az+bx}$

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

Each of the fraction is equal to $\dfrac{2}{a+b}$

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ We know that if a/b = c/d then, a/b = c/d = a + b/c+d

→ Using this identity on the LHS part we  get,

   $\dfrac{x+y}{ax+by} =\dfrac{y+z}{ay+bz} =\dfrac{z+x}{az+bx}=\dfrac{x+y+y+z+z+x}{ax+by+ay+bz+az+bx}$

→ Simplifying it we get,

  $\dfrac{x+y}{ax+by} =\dfrac{y+z}{ay+bz} =\dfrac{z+x}{az+bx}=\dfrac{2x+2y+2z}{ax+ay+az+bx+by+bz}$

→ Taking 2 common from the numerator and a and b common from the denominator we get,

  $\dfrac{x+y}{ax+by} =\dfrac{y+z}{ay+bz} =\dfrac{z+x}{az+bx}=\dfrac{2(x+y+z)}{a(x+y+z)+b(x+y+z)}$

→ Taking  x + y + z common from the denominator,

  $\dfrac{x+y}{ax+by} =\dfrac{y+z}{ay+bz} =\dfrac{z+x}{az+bx}=\dfrac{2(x+y+z)}{(x+y+z)(a+b)}$

→ Cancelling x + y + z on both numerator and denominator we get,

  $\dfrac{x+y}{ax+by} =\dfrac{y+z}{ay+bz} =\dfrac{z+x}{az+bx}=\dfrac{2}{a+b}$

= RHS

→ Hence proved.

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ If a/b = c/d then, a/b = c/d = a + b/c+d

Answer 2

If   , prove that each of the ratio is equal to , unless x + y + z = 0

Answer:

Step-by-step explanation:

Each of the fraction is equal to

→ We know that if a/b = c/d then, a/b = c/d = a + b/c+d

→ Using this identity on the LHS part we  get,

 

→ Simplifying it we get,

→ Taking 2 common from the numerator and a and b common from the denominator we get,

→ Taking  x + y + z common from the denominator,