Question

PLS solve question no. 20 . Urgent!!! [ONLY CORRECT ANSWER ACCEPTED NO NONSENSE OR ABSURD ANSWER OR U WILL BE REPORTED].

Answer 1

Given :-

(ax+by/a²+b²) = (by+cz/b²+c²) = (cz+ac/c²+a²)

To Prove :-

• (x/a) = (y/b) = (z/c)

Solution :-

Let us Assume That, (ax+by/a²+b²) = (by+cz/b²+c²) = (cz+ac/c²+a²) are Equal to k .( where k is any constant Number .)

Than,

→ (ax+by/a²+b²) = k

→ (ax + by) = k * (a² + b²) -------- Equation (1).

Similarly,

→ (by+cz/b²+c²) = k

→ (by + cz) = k * (b² + c²) -------- Equation (2).

And,

→ (cz+ac/c²+a²) = k

→ (cz+ax) = k * (c²+a²) -------- Equation (3).

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Now, Adding All Three Equations we get,

→ 2(ax + by + cz) = 2 * k * (a² + b² + c²)

→ (ax + by + cz) = k * (a² + b² + c²) ------ Equation (4).

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Now, Subtracting Equation (1) From Equation (4), we get,

→ (ax + by + cz) - (ax + by) = k * (a² + b² + c²) - k *(a² + b²)

→ ax - ax + by - by + cz = k[ a² + b² + c² - a² - b² ]

→ cz = k * c²

→ z = k * c

→ k = (z/c) . ----------- Equation (5) .

Similarly,

Subtracting Equation (2) From Equation (4), we get,

→ (ax + by + cz) - (cz + ac) = k * (a² + b² + c²) - k *(b² + c²)

→ ax + by + cz - cz = k[ a² + b² + c² - b² - c² ]

→ ax = k * a²

→ x = k * a

→ k = (x/a) . ---------- Equation (6) .

And,

Subtracting Equation (3) From Equation (4), we get,

→ (ax + by + cz) - (cz + ax) = k * (a² + b² + c²) - k *(c² + a²)

→ ax + by + cz - cz - ax = k[ a² + b² + c² - c² - a² ]

→ by = k * b²

→ y = k * b

→ k = (y/b) . ----------- Equation (7).

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Hence, Now, From Equation (5) , (6) & (7) , we can say That,

→ k = (z/c) = (x/a) = ( y/b)

Or,

→ (x/a) = (y/b) = (z/c) . (Proved).

Answer 2

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$\huge\tt{GIVEN:}$

• (ax+by/a²+b²) = (by+cz/vb²+c²) = (cz+ax/c²+a²)

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$\huge\tt{TO~PROVE}$

• x/y = y/b = z/c

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$\huge\tt{SOLUTION:}$

Let the sum of (ax+by/a²+b²) = (by+cz/vb²+c²) = (cz+ax/c²+a²) be k

↪(ax+by/a²+b²) = k

↪ (ax + by) = k ×(a² + b²) _____(EQ.1)

↪ (by+cz/b²+c²) = k

↪ (by + cz) = k ×(b² + c²) _____(EQ.2)

↪(cz+ac/c²+a²) = k

↪ (cz+ax) = k ×(c²+a²) ______(EQ.3)

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By adding the Equations We get,

↪ 2(ax + by + cz) = 2 ×k ×(a² + b² + c²)

↪(ax + by + cz) = k ×(a² + b² + c²) _______(EQ.4)

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Then,

We'll subtract equation 4 from 1

↪(ax + by + cz) - (ax + by) = k * (a² + b² + c²) - k ×(a² + b²)

↪ ax - ax + by - by + cz = k[ a² + b² + c² - a² - b² ]

↪ cz = k ×c²

↪ z = k ×c

↪k = (z/c) _____(EQ.5)

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Now,

Subtracting equation 4 from 2

↪(ax + by + cz) - (cz + ac) = k × (a² + b² + c²) - k ×(b² + c²)

↪ax + by + cz - cz = k[ a² + b² + c² - b² - c² ]

↪ax = k × a²

↪x = k ×a

↪ k = (x/a) ________(EQ.6)

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Again,

Subtracting equation 4 from 3

↪ (ax + by + cz) - (cz + ax) = k × (a² + b² + c²) - k ×(c² + a²)

↪ax + by + cz - cz - ax = k[ a² + b² + c² - c² - a² ]

↪by = k ×b²

↪ y = k×b

↪k = (y/b) ______(EQ.7)

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Now as we got our values in the equations 5,6 & 7, We can say that....

k = (z/c) = (x/a) = ( y/b)

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Therefore,

Proved that (x/y) = (y/b) = (z/c)

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