(b+c-a)x=(c+a-b)y=(a+b-c)z=(a+b+c)t then prove that 1/x+1/y+1/z=1/t
Answers
Question : -
(b+c-a)x = (c+a-b)y = (a+b-c)z = (a+b+c)t , then prove that (1)/(x)+(1)/(y)+(1)/(z) = (1)/(t)
ANSWER
Given : -
(b+c-a)x = (c+a-b)y = (a+b-c)z = (a+b+c)t
Required to prove : -
- (1)/(x)+(1)/(y)+(1)/(z) = (1)/(t)
Proof : -
Given that;
(b+c-a)x = (c+a-b)y = (a+b-c)z = (a+b+c)t
Now,
(b+c-a)x = (a+b+c)t
[(b+c-a)]/[(a+b+c)t] = (1)/(x) ....... (1)
Similarly,
(c+a-b)y = (a+b+c)t
[(c+a-b)]/[(a+b+c)t] = (1)/(y) ........(2)
Similarly,
(a+b-c)z = (a+b+c)t
[(a+b-c)]/[(a+b+c)t] = (1)/(z) ........(3)
Now,
Adding (1),(2)&(3)
(1)/(x) + (1)/(y) + (1)/(z)
»[(b+c-a)]/[(a+b+c)t]+[(c+a-b)]/[(a+b+c)t]+[(a+b-c)]/[(a+b+c)t]
»( [(b+c-a)]+[(c+a-b)]+[(a+b-c)] )/( [(a+b+c)t] )
since, all the denominator are equal ....
» (b+c-a+c+a-b+a+b-c)/( [a+b+c]t )
» ([a+b+c] x 1)/([a+b+c] x t)
» (1)/(t)
» RHS
Hence Proved !…
➨ We have to prove (1)/(x) + (1)/(y) + (1)/(z) = (1)/(t) and the given values are (b+c-a)x = (c+a-b)y = (a+b-c)z = (a+b+c)t
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According to the question now,
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➨ Equation 1
Now,
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➨ Equation 2
Now,
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➨ Equation 3
Done but now we have to add Equation 1 , 2 and 3. Let's add !
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Now, let's see what to do ( as we already see that the denominators be same ) so now let's add them !
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Let's cancel the digits.
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Henceforth, it's proved.