Solve simultaneously (1/u) + (1/v) = (1/3) .......... equation (i) (u^2 /v) + (v^2 /u) = 12 ........ equation (ii)


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Question Number 6824 by Tawakalitu. last updated on 30/Jul/16

$S o l v e s i m u l t a n e o u s l y$ $\frac{1}{u} + \frac{1}{v} = \frac{1}{3} . . . . . . . . . . e q u a t i o n (i)$ $\frac{u^{2}}{v} + \frac{v^{2}}{u} = 12 . . . . . . . . e q u a t i o n (i i)$
Commented by sou1618 last updated on 30/Jul/16
$(i)((v+u)/(uv))=(1/3)⇔ 3(u+v)=uv (ii)((u^3 +v^3 )/(uv))=12⇔(u+v)(u^2 −uv+v^2 )=12uv ⇔(u+v){(u+v)^2 −3uv}=12uv set { ((u+v=x)),((uv=y)) :} (i),(ii)⇒ { (((1). 3x=y)),(((2). x^3 −3xy=12y)) :} (2)⇒x^3 −3x(3x)=12(3x) ((1)) x^3 −9x^2 −36x=0 x(x−12)(x+3)=0 x=0,12,−3 (x,y)=(0,0) (12,36) (−3,−9) (u+v,uv)=(0,0) (12,36) (−3,−9) when u+v=0,uv=0 (i)is not defined (α+β,αβ)⇒z^2 −(α+β)z+αβ=0′s solution z^2 −12z+36=0⇒z=6 z^2 +3z−9=0⇒z=((−3±3(√5))/2) ⇒(u,v)=(6,6) (((−3±3(√5))/2),((−3∓3(√3))/2)) ∴(u,v)=(6,6) (((−3±3(√5))/2),((−3∓3(√3))/2)).$
$(i) \frac{v + u}{u v} = \frac{1}{3} \Leftrightarrow 3 (u + v) = u v$ $(i i) \frac{u^{3} + v^{3}}{u v} = 12 \Leftrightarrow (u + v) (u^{2} - u v + v^{2}) = 12 u v$ $\Leftrightarrow (u + v) {{(u + v)}^{2} - 3 u v} = 12 u v$ $s e t {\begin{cases} u + v = x \\ u v = y \end{cases}$ $(i), (i i) \Rightarrow {\begin{cases} (1) . 3 x = y \\ (2) . x^{3} - 3 x y = 12 y \end{cases}$ $(2) \Rightarrow x^{3} - 3 x (3 x) = 12 (3 x) ((1))$ $x^{3} - 9 x^{2} - 36 x = 0$ $x (x - 12) (x + 3) = 0$ $x = 0, 12, - 3$ $(x, y) = (0, 0) (12, 36) (- 3, - 9)$ $(u + v, u v) = (0, 0) (12, 36) (- 3, - 9)$ $w h e n u + v = 0, u v = 0$ $(i) i s n o t d e f i n e d$ $(α + β, α β) \Rightarrow z^{2} - (α + β) z + α β = 0^{'} s s o l u t i o n$ $z^{2} - 12 z + 36 = 0 \Rightarrow z = 6$ $z^{2} + 3 z - 9 = 0 \Rightarrow z = \frac{- 3 \pm 3 \sqrt{5}}{2}$ $\Rightarrow (u, v) = (6, 6) (\frac{- 3 \pm 3 \sqrt{5}}{2}, \frac{- 3 \mp 3 \sqrt{3}}{2})$ $∴ (u, v) = (6, 6) (\frac{- 3 \pm 3 \sqrt{5}}{2}, \frac{- 3 \mp 3 \sqrt{3}}{2}) .$
Commented by Tawakalitu. last updated on 30/Jul/16

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