If p = 4xy / (x+y) , prove that
(p + 2x / p - 2x ) + (p + 2y / p - 2y) = 2
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Here we have: p = 4xy/(x+y) which will imply that p/2x = 2y/(x+y). In other words we can say that:
(p+2x)/(p-2x) = (2y + x + y)/(2y – (x + y)) [By componendo and dividendo]
(p + 2x)/(p-2x) = (3y + x)/(y – x) ….. (i)
Again, we can say that p = 4xy/(x+y) and we can say that p/2y = 2x/(x + y)
Also, this means thta (p + 2y)/(p – 2y) = (2x + x + y)/(2x – (x +y)) [By componendo and dividendo]
Therefore, we have: (p + 2y)/(p – 2y) = (3x + y)/(x – y) ….(ii)
Adding (i) and (ii), we get:
(p + 2x)/(p – 2x) + (p + 2y)/(p – 2y) = (3y + x)/(y – x) + (3x + y)/(x – y)
In other words, this can be written as: (3y + x)/(y -x) – (3x + y)/(y – x) = (3y + x – 3x – y)/(y -x) = (2y – 2x)/(y – x)
Hence, we have: 2(y – x)/(y – x) = 2.
HENCE PROVED
(p+2x)/(p-2x) = (2y + x + y)/(2y – (x + y)) [By componendo and dividendo]
(p + 2x)/(p-2x) = (3y + x)/(y – x) ….. (i)
Again, we can say that p = 4xy/(x+y) and we can say that p/2y = 2x/(x + y)
Also, this means thta (p + 2y)/(p – 2y) = (2x + x + y)/(2x – (x +y)) [By componendo and dividendo]
Therefore, we have: (p + 2y)/(p – 2y) = (3x + y)/(x – y) ….(ii)
Adding (i) and (ii), we get:
(p + 2x)/(p – 2x) + (p + 2y)/(p – 2y) = (3y + x)/(y – x) + (3x + y)/(x – y)
In other words, this can be written as: (3y + x)/(y -x) – (3x + y)/(y – x) = (3y + x – 3x – y)/(y -x) = (2y – 2x)/(y – x)
Hence, we have: 2(y – x)/(y – x) = 2.
HENCE PROVED
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