1. If `a/b=c/d=e/f` then prove that `(2a+3c-4e)/(2b+3d-4f)=(5a-4c+3e)/(5b-4d+3f)`
Solution:
`a/b=c/d=e/f`
Suppose, `a/b=c/d=e/f=k` (say)
`:. a=bk,c=dk,e=fk`
Now LHS`=(2a+3c-4e)/(2b+3d-4f)`
`=(2bk+3dk-4fk)/(2b+3d-4f)`
`=(k(2b+3d-4f))/(2b+3d-4f)`
`"Now cancel the common factor "(2b+3d-4f)`
`=k`
Now RHS`=(5a-4c+3e)/(5b-4d+3f)`
`=(5bk-4dk+3fk)/(5b-4d+3f)`
`=(k(5b-4d+3f))/(5b-4d+3f)`
`"Now cancel the common factor "(5b-4d+3f)`
`=k`
2. If `x/(b^2-c^2)=y/(c^2-a^2)=z/(a^2-b^2)` then prove that `x+y+z=0`
Solution:
`x/(b^2-c^2)=y/(c^2-a^2)=z/(a^2-b^2)`
Suppose, `x/(b^2-c^2)=y/(c^2-a^2)=z/(a^2-b^2)=k` (say)
`:. x=b^2k-c^2k,y=c^2k-a^2k,z=a^2k-b^2k`
Now LHS`=x+y+z`
`=(b^2k-c^2k)+(c^2k-a^2k)+(a^2k-b^2k)`
`=0`
`=`RHS
This material is intended as a summary. Use your textbook for detail explanation.
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