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Find In ratio and proportion, if (5a+6b)/(7c)=(6b+7c)/(5a)=(7c+5a)/(6b) then prove that each ratio = 2,-1

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Your problem `->` In ratio and proportion, if (5a+6b)/(7c)=(6b+7c)/(5a)=(7c+5a)/(6b) then prove that each ratio = 2,-1


Here `(5a+6b)/(7c)=(6b+7c)/(5a)=(7c+5a)/(6b)`

Case-1 : If `5a+6b+7c!=0`, then

Each ratio`=(5a+6b+6b+7c+7c+5a)/(7c+5a+6b)`

`=(10a+12b+14c)/(7c+5a+6b)`

`=(2(5a+6b+7c))/(7c+5a+6b)`

Cancel the common factor `(5a+6b+7c)`

`=2`

Case-2 : If `5a+6b+7c=0`, then

`5a+6b=-7c`

Then, the first ratio `=(5a+6b)/(7c)`

`=(-7c)/(7c)`

Cancel the common factor `7c`

`=-1`

Hence, each ratio `=-1`.


Thus, the value of each ratio is `2` or `-1`.






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