1. If `x/(y+z)=y/(z+x)=z/(x+y)` then prove the value of each ratio is `1/2,-1`
Solution:
Here `x/(y+z)=y/(z+x)=z/(x+y)`
Case-1 : If `x+y+z!=0`, then
Each ratio`=(x+y+z)/(y+z+z+x+x+y)`
`=(x+y+z)/(2y+2z+2x)`
`=(x+y+z)/(2(y+z+x))`
Cancel the common factor `(x+y+z)`
`=(1)/(2)`
Case-2 : If `x+y+z=0`, then
`y+z=-x`
Then, the first ratio `=(x)/(y+z)`
`=(x)/(-x)`
Cancel the common factor `-x`
`=-1`
Hence, each ratio `=-1`.
Thus, the value of each ratio is `(1)/(2)` or `-1`.
2. If `(5a+6b)/(7c)=(6b+7c)/(5a)=(7c+5a)/(6b)` then prove the value of each ratio is `2,-1`
Solution:
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`.
This material is intended as a summary. Use your textbook for detail explanation.
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