Question

if a/x-2y+3z=b/y-2z+3x=c/z-2x+3y then show that each ratio is equal to a+b+c/2(x+y+z)​

Answer 1

Given :  a/(x-2y+3z)=b/(y-2z+3x)=c/(z-2x+3y)

To Find :  show that each ratio is equal to (a+b+c)/2(x+y+z)​

Solution:

a/(x-2y+3z)=b/(y-2z+3x)=c/(z-2x+3y) = K

=> a = K(x-2y+3z)

   b = K(y-2z+3x)

   c = K(z-2x+3y)

a + b + c =  K(x-2y+3z) +  K(y-2z+3x) +  K(z-2x+3y)

=> a + b + c = K ( 2x + 2y + 2z)

=> a + b + c  = K2(x + y + z)

=> (a + b + c)/2(x + y + z) = K

a/(x-2y+3z)=b/(y-2z+3x)=c/(z-2x+3y) = K

Hence a/(x-2y+3z)=b/(y-2z+3x)=c/(z-2x+3y) =  (a + b + c)/2(x + y + z)

QED

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Answer 2

Answer:

Step-by-step explanation: