{ ((x^2 +y^3 =23)),((x^3 +y^2 =32)) :} solve for x and y .


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Question Number 74801 by behi83417@gmail.com last updated on 30/Nov/19
${ ((x^2 +y^3 =23)),((x^3 +y^2 =32)) :} solve for x and y .$
${\begin{cases} x^{2} + y^{3} = 23 \\ x^{3} + y^{2} = 32 \end{cases} solve for x and y .$
Commented by MJS last updated on 30/Nov/19

$if you plot these you see {there}^{'} s only one$ $real solution$ $x \approx 2.96773$ $y \approx 2.42114$
Commented by MJS last updated on 01/Dec/19

$approximating I also found these$ $x \approx - 1.88268 \pm 2 . 70319 i; y \approx - 1.83861 \pm 2 . 44516 i$ $x \approx - 1.50880 \pm 3 . 04643 i; y \approx - 1.29910 \mp 2 . 87425 i$ $x \approx - 1.36613 \pm 2 . 49017 i; y \approx 3.03262 \pm . 247148 i$ $x \approx 3.27375 \pm . 139609 i; y \approx - 1.10548 \pm 2 . 02898 i$ $. . . so we get 9 solutions$
	Answered by behi83417@gmail.com last updated on 01/Dec/19

	$dear proph . MJS! thank you very much .$ $god bless you sir .$
	Answered by ajfour last updated on 01/Dec/19
	$(x^2 −23)^2 =(32−x^3 )^3 x^9 −96x^6 +x^4 +3072x^3 −46x^2 +23^2 −32^3 = 0 or x^2 (x−1)+y^2 (1−y)=9 =b−a x^2 (x+1)+y^2 (1+y)=55 =a+b ⇒ x^2 =(((b−a)(1+y)−(a+b)(1−y))/(2(xy−1))) y^2 =(((a+b)(x−1)−(b−a)(x+1))/(2(xy−1))) ⇒ (x^2 /y^2 )=((by−a)/(ax−b)) & x^2 y^2 (xy−1)=(by−a)(ax−b) let x=(b/a)u , y=(a/b)v ⇒ u^2 v^2 (uv−1)=ab(u−1)(v−1) (b^4 /a^4 )=(a/b)(((v−1)/(u−1))) ⇒ ((v−1)/(u−1))=(b^5 /a^5 ) = (k/1) ⇒ ((u+v−2)/(√(uv−(u+v)+1)))=((k+1)/(√k)) let u+v=s , uv=m ⇒ k(s−2)^2 =(k+1)^2 {(m−1)−(s−2)} & m^2 (m−1)=ab{(m−1)−(s−2)} ⇒ ab(1−((s−2)/(m−1)))=m^2 k(m−1)(1−(m^2 /(ab)))^2 =((m^2 (k+1)^2 )/(ab)) ...... ... (matter of quintic polynomial)$
	${(x^{2} - 23)}^{2} = {(32 - x^{3})}^{3}$ $x^{9} - 96 x^{6} + x^{4} + 3072 x^{3} - 46 x^{2}$ $+ 23^{2} - 32^{3} = 0$ $o r x^{2} (x - 1) + y^{2} (1 - y) = 9 = b - a$ $x^{2} (x + 1) + y^{2} (1 + y) = 55 = a + b$ $\Rightarrow x^{2} = \frac{(b - a) (1 + y) - (a + b) (1 - y)}{2 (x y - 1)}$ $y^{2} = \frac{(a + b) (x - 1) - (b - a) (x + 1)}{2 (x y - 1)}$ $\Rightarrow \frac{x^{2}}{y^{2}} = \frac{b y - a}{a x - b} &$ $x^{2} y^{2} (x y - 1) = (b y - a) (a x - b)$ $l e t x = \frac{b}{a} u, y = \frac{a}{b} v \Rightarrow$ $u^{2} v^{2} (u v - 1) = a b (u - 1) (v - 1)$ $\frac{b^{4}}{a^{4}} = \frac{a}{b} (\frac{v - 1}{u - 1})$ $\Rightarrow \frac{v - 1}{u - 1} = \frac{b^{5}}{a^{5}} = \frac{k}{1}$ $\Rightarrow \frac{u + v - 2}{\sqrt{u v - (u + v) + 1}} = \frac{k + 1}{\sqrt{k}}$ $l e t u + v = s, u v = m$ $\Rightarrow k {(s - 2)}^{2} = {(k + 1)}^{2} {(m - 1) - (s - 2)}$ $& m^{2} (m - 1) = a b {(m - 1) - (s - 2)}$ $\Rightarrow a b (1 - \frac{s - 2}{m - 1}) = m^{2}$ $k (m - 1) {(1 - \frac{m^{2}}{a b})}^{2} = \frac{m^{2} {(k + 1)}^{2}}{a b}$ $. . . . . .$ $. . . (m a t t e r o f q u i n t i c p o l y n o m i a l)$
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