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

Step-by-step explanation:

$(b+c-a)x=(c+a-b)y=(a+b-c)z=(a+b+c)t$

$(b+c-a)x=(a+b+c)t$

$x = \frac{(a + b + c)t}{(b + c - a)}$

$\frac{1}{x} = \frac{(b + c - a)}{(a + b + c)t}$

Similarly,

$\frac{1}{y} = \frac{(c + a - b)}{(a + b + c)t}$

$\frac{1}{z} = \frac{(a + b - c)}{(a + b + c)t}$

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$

$= \frac{(b + c - a)}{(a + b + c)t} + \frac{(c + a - b)}{(a + b + c)t} + \frac{(a + b - c)}{(a + b + c)t}$

$= \frac{b + c - a + c + a - b + a + b - c}{(a + b + c)t}$

$= \frac{a + b + c}{(a + b + c)t}$

$= \frac{1}{t}$

Therefore, $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{t}$

Answer 2

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}}}}

Giventhat

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

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

Toprove

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

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

Let

′

sprove

⚡

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

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

➨ {\bold{\bf{(1)/(x)}}}(1)/(x) Equation 1

Now,

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

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

➨ {\bold{\bf{(1)/(y)}}}(1)/(y) Equation 2

Now,

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

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

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

➨ {\bold{\bf{1 \times t}}}1×t

Henceforth, it's proved.