if ax+by=6, bx-ay=2 and x^2+y^2=4, then the value of (a^2+b^2) would be

• ax+by=6.....(1)
  bx-ay=2.....(2)
  squaring both equation and add them we will get
  x^2(a^2+b^2)+y^2(a^2+b^2)=40
  and (a^2+b^2)(x^2+y^2)=40
  now put d value of x^2+y^2
  then a^2+b^2=10 ans
• 12 years agoHelpfull: Yes(10) No(0)
• 10 sqar first tow eq an then add
• 12 years agoHelpfull: Yes(1) No(0)
• exam hall main hai kya agar nahy to questoins pls forward kar mera exam sunday ko bhai pls shyamleshd@gmail.com
• 12 years agoHelpfull: Yes(0) No(0)
• ax+by=6 =>a/y+b/x=6xy .....(1)
  bx-ay=2 =>b/y-a/x=2xy.......(2)
  adding 1+2 equations
  b{1/x+1/y}+a{1/y-1/x}=8xy
  b{y+x/xy}+a{x-y/xy}=8xy
  b[y+x]+a[x-y]=8x^2y^2
  y[b-a]+x[a+b]=8x^2y^2
  (a+b)[x-y]=8x^2y^2
  (a+b)^2{x^2+y^2-2xy}=64x^4y^4
  (a^2+b^2+2ab){4-2xy}=64x^4y^4
  a^2+b^2={(32x^4y^4)/(2-xy)-2ab}
• 12 years agoHelpfull: Yes(0) No(2)
• solve for a in eq.1 and eq.2 then used comparison method to solve for b
  (6-by)/x=(bx-2)/y
  6y-by^2=bx^2-2x => (6y+2x)/b=x^+y^2 => b= (3y+x)/2 then solve for b in eq.1 & eq.2 used comparison method to solve for a. a=(3x-y)/2
  so, [(3x-y)/2]^2+[(3y+x)/2]^2= 5(x^2+y^2)=20 therefore a^2+b^2=20
• 12 years agoHelpfull: Yes(0) No(4)