solve the follwing equation x(√(x )) + y(√(y )) = 3 and x(√(y )) + y(√(x )) = 2 someone solve the above equations in the following way x^3 + y^3 + 2xy(√(xy )) = 9.....(1) and x^2 y + y^2 x + 2xy(√(xy )) = 4......(2) (1) − (2) ⇒ (x − y)(x^2 − y^2 ) = 5 hence x = 3 and y = 2 which is obiviusly does not satisfy the original equations. where is the fallacy in the above solution? Please explain.


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Question Number 175801 by Lekhraj last updated on 07/Sep/22

$s o l v e t h e f o l l w i n g e q u a t i o n$ $x \sqrt{x} + y \sqrt{y} = 3 a n d x \sqrt{y} + y \sqrt{x} = 2$ $s o m e o n e s o l v e t h e a b o v e e q u a t i o n s i n t h e f o l l o w i n g w a y$ $x^{3} + y^{3} + 2 x y \sqrt{x y} = 9 . . . . . (1) a n d x^{2} y + y^{2} x + 2 x y \sqrt{x y} = 4 . . . . . . (2)$ $(1) - (2) \Rightarrow (x - y) (x^{2} - y^{2}) = 5$ $h e n c e x = 3 a n d y = 2 w h i c h i s o b i v i u s l y d o e s n o t s a t i s f y t h e$ $o r i g i n a l e q u a t i o n s .$ $w h e r e i s t h e f a l l a c y i n t h e a b o v e s o l u t i o n ? Please explain .$
Commented by mr W last updated on 07/Sep/22

$(x - y) (x^{2} - y^{2}) = 5$ $h a s i n f i n i t e s o l u t i o n s! n o t e v e r y$ $s o l u t i o n f r o m i t f u l f i l l s t h e o r i g i n a l$ $e q u a t i o n s . x = 3 a n d y = 2 i s o n l y o n e$ $p o s s i b l e s o l u t i o n o f (x - y) (x^{2} - y^{2}) = 5 .$
Commented by Frix last updated on 07/Sep/22

$squaring equations leads to false solutions$
Commented by Lekhraj last updated on 08/Sep/22

$Thank you .$
Commented by Lekhraj last updated on 08/Sep/22

$What about cubing equatins ?$
Commented by Frix last updated on 16/Sep/22

$cubing is also dangerous for complex numbers$
	Answered by behi834171 last updated on 07/Sep/22
	$x=t^2 ,y=s^2 ⇒ { ((t^3 +s^3 =3)),((st(s+t)=2)) :}⇒ { (((t+s)(t^2 +s^2 −ts)=3)),((ts(t+s)=2)) :} ⇒_(ts=q) ^(t+s=p) { ((p(p^2 −3q)=3)),((pq=2)) :}⇒p^3 −3pq=3⇒ ⇒p^3 −6=3⇒p^3 =9⇒p=t+s=(9)^(1/3) q=(2/p)=t.s=(2/( (9)^(1/3) )) ⇒z^2 −(9)^(1/3) .z+(2/( (9)^(1/3) ))=0 ⇒z_(1,2) =t or s =(((9)^(1/3) ±(√(3(3)^(1/3) −(8/( (9)^(1/3) )))))/2)⇒ t or s=((2.1±0.7)/2)=1.4 or 0.7 ⇒x=t^2 =(1.4)^2 or (0.7)^2 y=s^2 =(0.7)^2 or (1.4)^2 .■$
	$x = t^{2}, y = s^{2}$ $\Rightarrow {\begin{cases} t^{3} + s^{3} = 3 \\ s t (s + t) = 2 \end{cases} \Rightarrow {\begin{cases} (t + s) (t^{2} + s^{2} - t s) = 3 \\ t s (t + s) = 2 \end{cases}$ $\underset{t s = q}{\overset{t + s = p}{\Rightarrow}} {\begin{cases} p (p^{2} - 3 q) = 3 \\ p q = 2 \end{cases} \Rightarrow p^{3} - 3 p q = 3 \Rightarrow$ $\Rightarrow p^{3} - 6 = 3 \Rightarrow p^{3} = 9 \Rightarrow p = t + s = \sqrt[3]{9}$ $q = \frac{2}{p} = t . s = \frac{2}{\sqrt[3]{9}}$ $\Rightarrow z^{2} - \sqrt[3]{9} . z + \frac{2}{\sqrt[3]{9}} = 0$ $\Rightarrow z_{1, 2} = t o r s = \frac{\sqrt[3]{9} \pm \sqrt{3 \sqrt[3]{3} - \frac{8}{\sqrt[3]{9}}}}{2} \Rightarrow$ $t o r s = \frac{2.1 \pm 0.7}{2} = 1.4 o r 0.7$ $\Rightarrow x = t^{2} = {(1.4)}^{2} o r {(0.7)}^{2}$ $y = s^{2} = {(0.7)}^{2} o r {(1.4)}^{2} . ◼$
	Answered by Frix last updated on 07/Sep/22

	$x^{3 / 2} + y^{3 / 2} = 3$ ${x y}^{1 / 2} + x^{1 / 2} y = 2$ $x = u^{2} \land y = v^{2} \land u ⩾ 0 \land v ⩾ 0$ $u^{3} + v^{3} = 3$ $u^{2} v + {u v}^{2} = 2$ $(u^{3} + v^{3}) + 3 (u^{2} v + {u v}^{2}) = 3 + 3 \times 2$ ${(u + v)}^{3} = 9$ $v = 9^{1 / 3} - u$ $u^{3} + {(9^{1 / 3} - u)}^{3} = 3$ $\Rightarrow$ $u^{2} - 3^{2 / 3} u + \frac{2}{3^{2 / 3}} = 0$ $u = \frac{1}{3^{1 / 3}} \lor u = \frac{2}{3^{1 / 3}}$ $\Rightarrow$ $x = \frac{1}{3^{2 / 3}} \land y = \frac{4}{3^{2 / 3}} or the other way round$
	Commented by Lekhraj last updated on 08/Sep/22

	Commented by Lekhraj last updated on 08/Sep/22

	Commented by Lekhraj last updated on 08/Sep/22

	$Thank you very much .$
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