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Question Number 18864 by chernoaguero@gmail.com last updated on 31/Jul/17

Commented by mrW1 last updated on 31/Jul/17

$(x, y) = (1, 2), (2, 1), (- 2, 1), (1, - 2), (0, - 3), (- 3, 0)$
	Answered by behi.8.3.4.1.7@gmail.com last updated on 31/Jul/17
	$x+y=t,xy=s x^2 y^2 +x^2 +y^2 =9⇒s^2 +t^2 −2s=9 ⇒ { ((s^2 −2s+t^2 −9=0)),((t(s−1)=3⇒s−1=(3/t))) :} (s−1)^2 +t^2 −10=0⇒(9/t^2 )+t^2 −10=0 ⇒t^4 −10t^2 +9=0⇒ { ((t^2 =1⇒t=±1)),((t^2 =9⇒t=±3)) :} ⇒s= { ((1+(3/(±1))=4,−2)),((1+(3/(±3))=2,0)) :} 1) { ((x+y=1)),((xy=4)) :}⇒x+(4/x)=1⇒x^2 −x+4=0 ⇒x=((1±(√(1−16)))/2)=((1±i(√(15)))/2),y=((1∓i(√(15)))/2) 2) { ((x+y=−1)),((xy=−2)) :}⇒x−(2/x)=−1⇒x^2 +x−2=0 ⇒x=((−1±(√9))/2)=−2,1,y=1,−2 3) { ((x+y=3)),((xy=2)) :}⇒x+(2/x)=3⇒x^2 −3x+2=0 ⇒x=((3±1)/2)=2,1,y=1,2 4) { ((x+y=−3)),((xy=0⇒x=0,y=−3 ∨x=−3,y=0)) :} ⇒(x,y)= { ((((1±i(√(15)))/2),−2,1,2,0,−3)),((((1∓i(√(15)))/2),1,−2,2,−3,0 .■)) :}$
	$x + y = t, x y = s$ $x^{2} y^{2} + x^{2} + y^{2} = 9 \Rightarrow s^{2} + t^{2} - 2 s = 9$ $\Rightarrow {\begin{cases} s^{2} - 2 s + t^{2} - 9 = 0 \\ t (s - 1) = 3 \Rightarrow s - 1 = \frac{3}{t} \end{cases}$ ${(s - 1)}^{2} + t^{2} - 10 = 0 \Rightarrow \frac{9}{t^{2}} + t^{2} - 10 = 0$ $\Rightarrow t^{4} - 10 t^{2} + 9 = 0 \Rightarrow {\begin{cases} t^{2} = 1 \Rightarrow t = \pm 1 \\ t^{2} = 9 \Rightarrow t = \pm 3 \end{cases}$ $\Rightarrow s = {\begin{cases} 1 + \frac{3}{\pm 1} = 4, - 2 \\ 1 + \frac{3}{\pm 3} = 2, 0 \end{cases}$ $1) {\begin{cases} x + y = 1 \\ x y = 4 \end{cases} \Rightarrow x + \frac{4}{x} = 1 \Rightarrow x^{2} - x + 4 = 0$ $\Rightarrow x = \frac{1 \pm \sqrt{1 - 16}}{2} = \frac{1 \pm i \sqrt{15}}{2}, y = \frac{1 \mp i \sqrt{15}}{2}$ $2) {\begin{cases} x + y = - 1 \\ x y = - 2 \end{cases} \Rightarrow x - \frac{2}{x} = - 1 \Rightarrow x^{2} + x - 2 = 0$ $\Rightarrow x = \frac{- 1 \pm \sqrt{9}}{2} = - 2, 1, y = 1, - 2$ $3) {\begin{cases} x + y = 3 \\ x y = 2 \end{cases} \Rightarrow x + \frac{2}{x} = 3 \Rightarrow x^{2} - 3 x + 2 = 0$ $\Rightarrow x = \frac{3 \pm 1}{2} = 2, 1, y = 1, 2$ $4) {\begin{cases} x + y = - 3 \\ x y = 0 \Rightarrow x = 0, y = - 3 \lor x = - 3, y = 0 \end{cases}$ $\Rightarrow (x, y) = {\begin{cases} \frac{1 \pm i \sqrt{15}}{2}, - 2, 1, 2, 0, - 3 \\ \frac{1 \mp i \sqrt{15}}{2}, 1, - 2, 2, - 3, 0 . ◼ \end{cases}$
	Commented by chernoaguero@gmail.com last updated on 31/Jul/17

	$t h a n k z s i r$
	Answered by mrW1 last updated on 31/Jul/17

	$let u = x + y$ $let v = xy - 1$ $(x^{2} + 1) (y^{2} + 1) = 10$ $x^{2} + y^{2} + x^{2} y^{2} + 1 = 10$ $x^{2} + y^{2} + 2 xy + x^{2} y^{2} - 2 xy + 1 = 10$ ${(x + y)}^{2} + {(xy - 1)}^{2} = 10$ $u^{2} + v^{2} = 10 . . . (i)$ $(x + y) (xy - 1) = 3$ $uv = 3 . . . (ii)$ $(i) :$ $u^{2} + v^{2} + 2 uv - 2 uv = 10$ $\Rightarrow {(u + v)}^{2} = 6 + 10 = 16$ $\Rightarrow u + v = \pm 4$ $u^{2} + v^{2} - 2 uv + 2 uv = 10$ $\Rightarrow {(u - v)}^{2} = - 6 + 10 = 4$ $\Rightarrow u - v = \pm 2$ $\Rightarrow u = \frac{\pm 4 \pm 2}{2} = 3, 1, - 1, - 3$ $\Rightarrow v = \frac{\pm 4 \mp 2}{2} = 1, 3, - 3, - 1$ $with (u, v) = (3, 1) :$ $x + y = 3$ $xy - 1 = 1 \Rightarrow xy = 2$ $x^{2} + y^{2} + 2 xy = 9$ $x^{2} + y^{2} - 2 xy = 9 - 4 xy = 9 - 8 = 1$ ${(x - y)}^{2} = 1$ $x - y = \pm 1$ $\Rightarrow x = \frac{3 \pm 1}{2} = 2, 1$ $\Rightarrow y = \frac{3 \mp 1}{2} = 1, 2$ $with (u, v) = (1, 3) :$ $x + y = 1$ $xy - 1 = 3 \Rightarrow xy = 4$ ${(x - y)}^{2} = 1 - 16 = - 15$ $\Rightarrow no real solution .$ $complex solution :$ $x - y = \pm \sqrt{15} i$ $\Rightarrow x = \frac{1 \pm \sqrt{15} i}{2}$ $\Rightarrow y = \frac{1 \mp \sqrt{15} i}{2}$ $with (u, v) = (- 1, - 3) :$ $x + y = - 1$ $xy - 1 = - 3 \Rightarrow xy = - 2$ ${(x - y)}^{2} = 1 + 8 = 9$ $\Rightarrow x - y = \pm 3$ $\Rightarrow x = \frac{- 1 \pm 3}{2} = 1, - 2$ $\Rightarrow y = \frac{- 1 \mp 3}{2} = - 2, 1$ $with (u, v) = (- 3, - 1) :$ $x + y = - 3$ $xy - 1 = - 1 \Rightarrow xy = 0$ $\Rightarrow x = 0, - 3$ $\Rightarrow y = - 3, 0$ $\Rightarrow all real solutions :$ $(x, y) = (2, 1), (1, 2), (1, - 2), (- 2, 1), (0, - 3), (- 3, 0)$ $\Rightarrow all complex solutions :$ $(x, y) = (\frac{1 + \sqrt{15} i}{2}, \frac{1 - \sqrt{15} i}{2}), (\frac{1 - \sqrt{15} i}{2}, \frac{1 + \sqrt{15} i}{2})$
	Commented by chernoaguero@gmail.com last updated on 31/Jul/17

	$T h a n k z s i r$
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