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Given, (x+y) / (3a-b) = (y+z) / (3b-c) = (z+x) / (3c-a). Show that (x+y+z) / (a+b+c) = (ax+by+cx) / (a²+b²+c²).

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10 months ago

```							Step-by-step explanation:( x + y ) / ( 3a - b ) .........( 1 )( y + z ) / ( 3b - c )  .........( 2 )( z + x ) / ( 3c - a ) .........( 3 )Add equation 1, 2 and 3( 2x + 2y + 2z ) / (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 )= ( x / a ) = ( y / b ) = ( z / c )⇒ ax / a²  =  by / b² = cz / c²= ( ax + by + cz ) / ( a² + b² + c² )It proves ( x+ y + z ) / ( a + b +c ) = ( ax + by + cz ) / ( a² + b² + c² )
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10 months ago
```							I cannot understand the following step: " ( x+ y + z ) / ( a + b +c )= ( x / a ) = ( y / b ) = ( z / c )" Please show the full step without any jumps.
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10 months ago
```							hello sayanta a much better way to solve is:let (x+y) / (3a-b) = (y+z) / (3b-c) = (z+x) / (3c-a) = kby componendo and dividendo, we havek= [(x+y) + (y+z) + (z+x)]/[(3a – b) + (3b – c) + (3c – a)]or k= (x+y+z)/(a+b+c).......(*)also x+y= k(3a – b)....(1)y+z= k(3b – c)......(2)and z+x= k(3c – a).......(3)solving these 3 eqns we havex= k(a+b+c) – k(3b – c)= k(a – 2b + 2c)y= k(b – 2c + 2a)and z= k(c – 2a + 2b)so, ax+by+cz= k(a^2 – 2ab + 2ac + b^2 – 2bc + 2ab + c^2 – 2ac + 2bc)= k(a^2 + b^2 + c^2)or (ax+by+cx) / (a²+b²+c²) = k.......(#)from (*) and (#), we havek= (x+y+z) / (a+b+c) = (ax+by+cx) / (a²+b²+c²). kindly approve :))
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10 months ago
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