{ (((1/x)+(1/y)=34)),(((1/(√x))+(1/(√y))=23−(1/(√(xy))) )) :} find the solution.


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Question Number 80702 by john santu last updated on 05/Feb/20

${\begin{cases} \frac{1}{x} + \frac{1}{y} = 34 \\ \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 23 - \frac{1}{\sqrt{x y}} \end{cases}$ $f i n d t h e s o l u t i o n .$
Commented by mind is power last updated on 05/Feb/20
$let (1/(√x))+(1/(√y))=u (1/(√(xy))). ⇒u^2 −2v=34 u=23−v ⇒(23−v)^2 −2v=34 v^2 −48v+495=0 v∈{15,28} u≥0⇒v=15 ⇒u=8 { (((1/(√x))+(1/(√y))=8)),(((1/(√x)).(1/(√y))=15)) :} (1/(√x)),(1/(√y)) root of X^2 −8X+15 X∈{3,5} ⇒(x,y)∈{((1/9),(1/(25)));((1/(25)),(1/9))}$
$l e t \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = u$ $\frac{1}{\sqrt{x y}} .$ $\Rightarrow u^{2} - 2 v = 34$ $u = 23 - v$ $\Rightarrow {(23 - v)}^{2} - 2 v = 34$ $v^{2} - 48 v + 495 = 0$ $v \in {15, 28}$ $u ⩾ 0 \Rightarrow v = 15 \Rightarrow u = 8$ ${\begin{cases} \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 8 \\ \frac{1}{\sqrt{x}} . \frac{1}{\sqrt{y}} = 15 \end{cases}$ $\frac{1}{\sqrt{x}}, \frac{1}{\sqrt{y}} r o o t o f X^{2} - 8 X + 15$ $X \in {3, 5}$ $\Rightarrow (x, y) \in {(\frac{1}{9}, \frac{1}{25}); (\frac{1}{25}, \frac{1}{9})}$
Commented by john santu last updated on 05/Feb/20
$let (1/(√x))+(1/(√y))=u, (1/(√(xy)))=v u^2 −2v=34 , u+v=23 u^2 +2u−80=0 ⇒ { ((u=8 )),((u=−10)) :} for u = 8 , v=15 (1/(√x))+(1/(√y))=8 ∧(1/(√(xy)))=15 (1/(√x))=3 , (1/(√x)) =5 ⇒ x = (1/9) , x=(1/(25)) ∴ ((1/9),(1/(25))) , ((1/(25)), (1/9))$
$l e t \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = u, \frac{1}{\sqrt{x y}} = v$ $u^{2} - 2 v = 34, u + v = 23$ $u^{2} + 2 u - 80 = 0 \Rightarrow {\begin{cases} u = 8 \\ u = - 10 \end{cases}$ $f o r u = 8, v = 15$ $\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 8 \land \frac{1}{\sqrt{x y}} = 15$ $\frac{1}{\sqrt{x}} = 3, \frac{1}{\sqrt{x}} = 5 \Rightarrow x = \frac{1}{9}, x = \frac{1}{25}$ $∴ (\frac{1}{9}, \frac{1}{25}), (\frac{1}{25}, \frac{1}{9})$
Commented by john santu last updated on 05/Feb/20

$g r e a t s i r$
Commented by MJS last updated on 06/Feb/20
$(1/(√x))=u−v; (1/(√y))=u+v; u−v>0∧u+v>0 { ((v^2 =17−u^2 )),((u^2 +u−20=0)) :} { ((v_1 =±2(√2)i; v_2 =±1)),((u_1 =−5; u_2 =4)) :} [u_1 , v_1 not useable] (1/(√x))=5∧(1/(√y))=3 ∨ (1/(√x))=3∧(1/(√y))=5 x=(1/(25))∧y=(1/9) ∨ x=(1/9)∧y=(1/(25))$
$\frac{1}{\sqrt{x}} = u - v; \frac{1}{\sqrt{y}} = u + v; u - v > 0 \land u + v > 0$ ${\begin{cases} v^{2} = 17 - u^{2} \\ u^{2} + u - 20 = 0 \end{cases}$ ${\begin{cases} v_{1} = \pm 2 \sqrt{2} i; v_{2} = \pm 1 \\ u_{1} = - 5; u_{2} = 4 \end{cases} [u_{1}, v_{1} not useable]$ $\frac{1}{\sqrt{x}} = 5 \land \frac{1}{\sqrt{y}} = 3 \lor \frac{1}{\sqrt{x}} = 3 \land \frac{1}{\sqrt{y}} = 5$ $x = \frac{1}{25} \land y = \frac{1}{9} \lor x = \frac{1}{9} \land y = \frac{1}{25}$
	Answered by behi83417@gmail.com last updated on 05/Feb/20
	$(1/x)=t^2 ,(1/y)=s^2 ⇒ { ((t^2 +s^2 =34)),((t+s=23−ts)) :}⇒_(ts=q) ^(t+s=p) { (((t+s)^2 −2ts=34)),((t+s+ts=23)) :} ⇒ { ((p^2 −2q=34)),((p+q=23)) :}⇒p^2 −2(23−p)−34=0 ⇒p^2 +2p+1=34+46+1⇒(p+1)^2 =81 ⇒p+1=±9⇒ { ((p=8⇒q=15)),((p=−10⇒q=33)) :} ⇒ { (([p=8,q=15]⇒[(1/(√x))=3,5,(1/(√y))=5,3])),((⇒(x,y)=((1/9),(1/(25))),((1/(25)),(1/9)))) :} ⇒ { (([p=−10,q=33]⇒z^2 +10z+33=0)),((z=−5±(√(25−4×33))=−5±(√(107))i)) :} ⇒(1/(√x))=−5±(√(107))i⇒ { ((x=(1/((−5−(√(107))i)^2 )))),((y=(1/((−5+(√(107))i)^2 )))) :}$
	$\frac{1}{x} = t^{2}, \frac{1}{y} = s^{2}$ $\Rightarrow {\begin{cases} t^{2} + s^{2} = 34 \\ t + s = 23 - ts \end{cases} \underset{ts = q}{\overset{t + s = p}{\Rightarrow}} {\begin{cases} {(t + s)}^{2} - 2 ts = 34 \\ t + s + ts = 23 \end{cases}$ $\Rightarrow {\begin{cases} p^{2} - 2 q = 34 \\ p + q = 23 \end{cases} \Rightarrow p^{2} - 2 (23 - p) - 34 = 0$ $\Rightarrow p^{2} + 2 p + 1 = 34 + 46 + 1 \Rightarrow {(p + 1)}^{2} = 81$ $\Rightarrow p + 1 = \pm 9 \Rightarrow {\begin{cases} p = 8 \Rightarrow q = 15 \\ p = - 10 \Rightarrow q = 33 \end{cases}$ $\Rightarrow {\begin{cases} [p = 8, q = 15] \Rightarrow [\frac{1}{\sqrt{x}} = 3, 5, \frac{1}{\sqrt{y}} = 5, 3] \\ \Rightarrow (x, y) = (\frac{1}{9}, \frac{1}{25}), (\frac{1}{25}, \frac{1}{9}) \end{cases}$ $\Rightarrow {\begin{cases} [p = - 10, q = 33] \Rightarrow z^{2} + 10 z + 33 = 0 \\ z = - 5 \pm \sqrt{25 - 4 \times 33} = - 5 \pm \sqrt{107} i \end{cases}$ $\Rightarrow \frac{1}{\sqrt{x}} = - 5 \pm \sqrt{107} i \Rightarrow {\begin{cases} x = \frac{1}{{(- 5 - \sqrt{107} i)}^{2}} \\ y = \frac{1}{{(- 5 + \sqrt{107} i)}^{2}} \end{cases}$
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