Question

If x+y/3a-b=y+z/3b-c=z+x/3c-a , then show that x+y+z/a+b+c=ax+by+cz/a^2+b^2+c^2.​

Answer 1

||✪✪ QUESTION ✪✪||

If x+y/3a-b=y+z/3b-c=z+x/3c-a , then show that x+y+z/a+b+c=ax+by+cz/a^2+b^2+c^2. ?

|| ✰✰ ANSWER ✰✰ ||

Given That :-

→ (x+y/3a-b) = (y+z/3b-c) = (z+x/3c-a)

Now, According to Ratio and proportion :-

If a/x = b/y = c/z , Than , they are also Equal to , (a + b + c)/(x + y + z) .

So,

→ (x+y/3a-b) = (y+z/3b-c) = (z+x/3c-a)

→ ( x + y + y + z + z + x) / (3a - b + 3b - c + 3c - a)

→ (2x + 2y + 2z) / ( 2a + 2b + 2c)

→ 2(x + y + z) /2(a + b + c)

→ (x + y + z) /(a + b + c) --------- Equation (1)

Again, Using Same concept in reverse, we can say That ,

→ (x/a) + (y/b) + (z/c)

Now, Multiply and divide Each Term by a , b and c ,

→ (ax/a²) + (by/b²) + (cz/c²)

Using same Ratio and proportional ,

→ (ax+by+cz) / (a² + b² + c²) --------- Equation (2)

Hence, From Equation (1) & (2) ,

→ (x + y + z) /(a + b + c) = (ax+by+cz) / (a² + b² + c²)

✪✪ Proved ✪✪

Answer 2

Given:

★ x+y/3a-b=y+z/3b-c=z+x/3c-a

$\rule{200}2$

To Prove:

★ x+y+z/a+b+c= ax+by+cz/a²+b²+c²

$\rule{200}2$

Proof:

★ Applying rule of identity

a1/b1=a2/b2=a3/b3

→ (a1+a2+a3)/(b1+b2+b3) .....(1)

→(a1-a2+a3)/(b1-b3+b1).......(2)

→(a2-a3+a1)/(b2-b3+b1).....(3)

→(a3-a1+a2)/(b3-b1+b2).....(4)

★ Now, apply to the given equation

x+y/3a-b=y+z/3b-c=z+x/3c-a

→ x+y+y+z+z+x/3a-b+3b-3c-a....using (1)

=x+y+z/a+b+c

→x+y-y-z+z+x/3a-b-3b-c+3c-a.....by(2)

→2x/2a-4b+4c

→x/a-2b+2c

→ax/a²-2ab+2ac .....result A

→y+z-z-x+x+y/3b-c-3c+3a-b.....by(3)

→2y/2b-4c+4a

→y/b-2c+2a

→by/b²-2bc+2ab.....result B

→z+x-x-y+y+z/3c-a+b+3b-c

→z/c-2a-2b

=cz/c²-2ca+2bc.....result C

★Now adding result A, B and C;

= ax+by+cz/(a²-2ab+2ca)+(b²-2bc+2ab)+(c²-2ca+2bc)

= ax+by+cz/a²+b²+c²

$\rule{200}2$