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If-3a-b-x-y-3b-c-y-z-3c-a-z-x-then-show-that-a-b-c-x-y-z-a-2-b-2-c-2-ax-by-cz-

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Question Number 202290 by MATHEMATICSAM last updated on 24/Dec/23
If ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) then show that ((a + b + c)/(x + y + z)) = ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) .
If3a−bx+y=3b−cy+z=3c−az+xthenshowthata+b+cx+y+z=a2+b2+c2ax+by+cz.
Answered by Rasheed.Sindhi last updated on 24/Dec/23
If ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) then show that ((a + b + c)/(x + y + z)) = ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) ⇒((x+y)/(3a − b))=((y+z)/(3b − c))=((z+x)/(3c − a))=k (say) x+y=k(3a − b)...(i) y+z=k(3b − c)...(ii) z+x=k(3c − a)...(iii) (i)+(ii)+(iii): 2(x+y+z)=k{(3a − b)+(3b − c)+(3c − a)} x+y+z=k(a+b+c)...(iv) (iv)−(ii): x=k(a+b+c)−k(3b − c) =k(a−2b+2c) (iv)−(iii): y=k(a+b+c)−k(3c − a) =k(2a+b−2c) (iv)−(i): z=k(a+b+c)−k(3a − b) =k(−2a+2b+c) To prove: ((a + b + c)/(x + y + z)) = ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) lhs: ((a + b + c)/(x + y + z))=((a + b + c)/(k(a+b+c)))=(1/k) rhs: ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) = ((a^(2 ) + b^2 + c^2 )/(a(k(a−2b+2c)) + b(k(2a+b−2c)) + c(k(−2a+2b+c)))) = ((a^(2 ) + b^2 + c^2 )/(k{a(a−2b+2c)) + b(2a+b−2c)) + c(−2a+2b+c))})) = ((a^(2 ) + b^2 + c^2 )/(k{a^2 −2ab+2ca + 2ab+b^2 −2bc −2ca+2bc+c^2 })) = ((a^(2 ) + b^2 + c^2 )/(k{a^2 +b^2 +c^2 }))=(1/k) ∵ lhs=rhs=1/k ∴ Proved
If3a−bx+y=3b−cy+z=3c−az+xthenshowthata+b+cx+y+z=a2+b2+c2ax+by+cz3a−bx+y=3b−cy+z=3c−az+x⇒x+y3a−b=y+z3b−c=z+x3c−a=k(say)x+y=k(3a−b)…(i)y+z=k(3b−c)…(ii)z+x=k(3c−a)…(iii)(i)+(ii)+(iii):2(x+y+z)=k{(3a−b)+(3b−c)+(3c−a)}x+y+z=k(a+b+c)…(iv)(iv)−(ii):x=k(a+b+c)−k(3b−c)=k(a−2b+2c)(iv)−(iii):y=k(a+b+c)−k(3c−a)=k(2a+b−2c)(iv)−(i):z=k(a+b+c)−k(3a−b)=k(−2a+2b+c)Toprove:a+b+cx+y+z=a2+b2+c2ax+by+czlhs:a+b+cx+y+z=a+b+ck(a+b+c)=1krhs:a2+b2+c2ax+by+cz=a2+b2+c2a(k(a−2b+2c))+b(k(2a+b−2c))+c(k(−2a+2b+c))=a2+b2+c2k{a(a−2b+2c))+b(2a+b−2c))+c(−2a+2b+c))}=a2+b2+c2k{a2−2ab+2ca+2ab+b2−2bc−2ca+2bc+c2}=a2+b2+c2k{a2+b2+c2}=1k∵lhs=rhs=1/k∴Proved
Answered by som(math1967) last updated on 24/Dec/23
Each ratio =((3a−b+3b−c+3c−a)/(x+y+y+z+z+x)) =((2(a+b+c))/(2(x+y+z)))=((a+b+c)/(x+y+z)) ⇒((a+b+c)/(x+y+z))=((a+b+c−3a+b)/(x+y+z−x−y)) =((a+b+c−3b+c)/(x+y+z−y−z))=((a+b+c−3c+a)/(x+y+z−z−x)) ⇒((a+b+c)/(x+y+z))=((2b−2a+c)/z)=((a−2b+2c)/x) =((2a−2c+b)/y) ⇒((a+b+c)/(x+y+z))=((2bc−2ac+c^2 )/(cz))=((a^2 −2ab+2ac)/(ax)) =((2ab−2bc+b^2 )/(by)) ⇒((a+b+c)/(x+y+z)) =((2bc−2ac+c^2 +a^2 −2ab+2ac+2ab−2bc+b^2 )/(cz+ax+by)) ⇒((a+b+c)/(x+y+z))=((a^2 +b^2 +c^2 )/(ax+by+cz))
Eachratio=3a−b+3b−c+3c−ax+y+y+z+z+x=2(a+b+c)2(x+y+z)=a+b+cx+y+z⇒a+b+cx+y+z=a+b+c−3a+bx+y+z−x−y=a+b+c−3b+cx+y+z−y−z=a+b+c−3c+ax+y+z−z−x⇒a+b+cx+y+z=2b−2a+cz=a−2b+2cx=2a−2c+by⇒a+b+cx+y+z=2bc−2ac+c2cz=a2−2ab+2acax=2ab−2bc+b2by⇒a+b+cx+y+z=2bc−2ac+c2+a2−2ab+2ac+2ab−2bc+b2cz+ax+by⇒a+b+cx+y+z=a2+b2+c2ax+by+cz

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