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 1

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 !…

Answer 2

${\bold{\sf{\underline{Question}}}}$

➨ 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

${\bold{\sf{\underline{Given \: that}}}}$

➨ ${\bold{\bf{(b+c-a)x = (c+a-b)y = (a+b-c)z = (a+b+c)t}}}$

${\bold{\sf{\underline{To \: prove}}}}$

➨ ${\bold{\bf{(1)/(x) + (1)/(y) + (1)/(z) = (1)/(t)}}}$

${\bold{\sf{\underline{Let's \: prove}}}}$ ⚡

➨ ${\bold{\bf{(b+c-a)x = (c+a-b)y = (a+b-c)z = (a+b+c)t}}}$

According to the question now,

➨ ${\bold{\bf{(b+c-a)x = (a+b+c)t}}}$

➨ ${\bold{\bf{[(b+c-a)]/[a+b+c)t]}}}$

➨ ${\bold{\bf{(1)/(x)}}}$ Equation 1

Now,

➨ ${\bold{\bf{(c+a-b)]/[a+b+c)t]}}}$

➨ ${\bold{\bf{(c+a-b)]/[(a+b+c)t]}}}$

➨ ${\bold{\bf{(1)/(y)}}}$ Equation 2

Now,

➨ ${\bold{\bf{(a+b-c)z=(a+b+c)t}}}$

➨ ${\bold{\bf{[(a+b+c)]/[(a+b+c)t]}}}$

➨ ${\bold{\bf{(1)/(z)}}}$ Equation 3

Done but now we have to add Equation 1 , 2 and 3. Let's add !

➨ ${\bold{\bf{(1)/(x) + (1)/(y) + (1)/(z)}}}$

➨ ${\bold{\bf{[(b+c-a)] / [(a+b+c)t] + [(c+a-b)] / [(a+b+c)t] + [(a+-c)] / [(a+b+c)t]}}}$

➨ ${\bold{\bf{[(b+c-a)] + [(c+a-b)] + [(a+b-c)] / [(a+b+c)t]}}}$

Now, let's see what to do ( as we already see that the denominators be same ) so now let's add them !

➨ ${\bold{\bf{b+c-a+c+a-b+a+b-c) / [(a+b+c)t]}}}$

Let's cancel the digits.

➨ ${\bold{\bf{[(a+b+c) \times 1] / [a+b+c] \times t}}}$

➨ ${\bold{\bf{1 \times t}}}$

Henceforth, it's proved.