Question

Que-3 if 2y + 2z-x/a
=2z+2x-y/b=2x+2y-z/c then
prove that 9x/2b+2c-a
=9y/2c+2a-b=9z/2a+2b-c​

Answer 1

Answer:

b=(2x+2y-2z)/c it should be b=(2x+2y-z)/c ,

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

a / (2y+2z-x) = b/(2z+2x-y) = c/(2x+2y-z) = k (constant)

(2b + 2c-a) = k (4z+4x-2y + 4x +4y - 2z - 2y-2z +x) = 9 kx

(2c+2a-b)= k( 4x+4y - 2z + 4y+4z - 2x- 2z -2x +y) = 9 ky

similarly it can be proved

(2a+2b-c)= 9 kz

Hence

x/(2b+2c-a)=y/(2c+2a-b)=z/(2a+2b-c) = 1/ 9k

Step-by-step explanation:

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Answer 2

Answer:

2y+2z−xa=2z+2x−yb=2x+2y−zc

=> 4y+4z−2x2a=4z+4x−2y2b=4x+4y−2z2c=2y+2z−xa=2z+2x−yb=2x+2y−zc

=> (4y+4z−2x)+(4z+4x−2y)2a+2b=(4z+4x−2y)+(4x+4y−2z)2b+2c=(4x+4y−2z)+(4y+4z−2x)2c+2a=2y+2z−xa=2z+2x−yb=2x+2y−zc

=> 8x+2y+2z2b+2c=2x+8y+2z2c+2a=2x+2y+8z2a+2b=−(2y+2z−x)−a=−(2z+2x−y)−b=−(2x+2y−z)−c

Applying addendo between 1st and the 4th ratio, between 2nd and 5th ratio and between 3rd and 6th ration in pairs we get

=>  9x2b+2c−a=9y2c+2a−b=9z2a+2b−c

Cancelling out 9 on the numerator and by taking the reciprocal of each of the ratios, we get

2b+2c−ax=2c+2a−by=2a+2b−cz

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Step-by-step explanation: