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Question Number 181902 by Acem last updated on 01/Dec/22

	Answered by Rasheed.Sindhi last updated on 02/Dec/22
	$(x^2 +xy+y^2 )(√(x^2 +y^2 )) =185....(i) (x^2 −xy+y^2 )(√(x^2 +y^2 )) =65.....(ii) [(√(x^2 +y^2 )) ≠0] (i)/(ii): ((x^2 +xy+y^2 )/(x^2 −xy+y^2 ))=((185)/(65))=((37)/(13)) ((x^2 +xy+y^2 −x^2 +xy−y^2 )/(x^2 +xy+y^2 +x^2 −xy+y^2 ))=((37−13)/(37+13)) [Componendo-Dividendo property] ((2xy)/(2x^2 +2y^2 ))=((37−13)/(37+13)) ((xy)/(x^2 +y^2 ))=((12)/(25)) 12x^2 −25xy+12y^2 =0 ((12x)/(25y))+((12y)/(25x))=1 ((12)/(25))(t+(1/t))=1 t+(1/t)=((25)/(12)) 12t^2 −25t+12=0 t=((25±(√(625−576)))/(24))=((25±7)/(24))=(4/3),(3/4) (x/y)=(4/3),(3/4) 3x=4y ∨ 4x=3y y=((3x)/4),((4x)/3) y=((3x)/4): (i)⇒( x^2 +x(((3x)/4))+(((3x)/4))^2 )(√(x^2 +(((3x)/4))^2 )) =185 ((16x^2 +12x^2 +9x^2 )/(16))(√((16x^2 +9x^2 )/(16))) =185 ((37x^2 )/(16))∙((5∣x∣)/4) =185 ∣x∣^3 =((185∙64)/(5∙37))=4^3 ⇒∣x∣=4⇒x=±4 ⇒y=((3(±4))/4)=±3 y=((4x)/3): (i)⇒(x^2 +x(((4x)/3))+(((4x)/3))^2 )(√(x^2 +(((4x)/3))^2 )) =185 ((9x^2 +12x^2 +16x^2 )/9)(√((9x^2 +16x^2 )/9)) =185 ((37x^2 )/9)∙((5∣x∣)/3)=185 ∣x∣^3 =((185×27)/(37×5))⇒∣x∣=3⇒x=±3 ⇒y=((4(±3))/3)=±4 (x,y)={(4,3),(3,4),(−4,−3),(−3,−4)}$
	$(x^{2} + x y + y^{2}) \sqrt{x^{2} + y^{2}} = 185 . . . . (i)$ $(x^{2} - x y + y^{2}) \sqrt{x^{2} + y^{2}} = 65 . . . . . (i i)$ $[\sqrt{x^{2} + y^{2}} \neq 0]$ $(i) / (i i) : \frac{x^{2} + x y + y^{2}}{x^{2} - x y + y^{2}} = \frac{185}{65} = \frac{37}{13}$ $\frac{x^{2} + x y + y^{2} - x^{2} + x y - y^{2}}{x^{2} + x y + y^{2} + x^{2} - x y + y^{2}} = \frac{37 - 13}{37 + 13}$ $[C o m p o n e n d o - D i v i d e n d o p r o p e r t y]$ $\frac{2 x y}{2 x^{2} + 2 y^{2}} = \frac{37 - 13}{37 + 13}$ $\frac{x y}{x^{2} + y^{2}} = \frac{12}{25}$ $12 x^{2} - 25 x y + 12 y^{2} = 0$ $\frac{12 x}{25 y} + \frac{12 y}{25 x} = 1$ $\frac{12}{25} (t + \frac{1}{t}) = 1$ $t + \frac{1}{t} = \frac{25}{12}$ $12 t^{2} - 25 t + 12 = 0$ $t = \frac{25 \pm \sqrt{625 - 576}}{24} = \frac{25 \pm 7}{24} = \frac{4}{3}, \frac{3}{4}$ $\frac{x}{y} = \frac{4}{3}, \frac{3}{4}$ $3 x = 4 y \lor 4 x = 3 y$ $y = \frac{3 x}{4}, \frac{4 x}{3}$ $y = \frac{3 x}{4} :$ $(i) \Rightarrow (x^{2} + x (\frac{3 x}{4}) + {(\frac{3 x}{4})}^{2}) \sqrt{x^{2} + {(\frac{3 x}{4})}^{2}} = 185$ $\frac{16 x^{2} + 12 x^{2} + 9 x^{2}}{16} \sqrt{\frac{16 x^{2} + 9 x^{2}}{16}} = 185$ $\frac{37 x^{2}}{16} \cdot \frac{5 ∣ x ∣}{4} = 185$ $∣ x ∣^{3} = \frac{185 \cdot 64}{5 \cdot 37} = 4^{3} \Rightarrow∣ x ∣= 4 \Rightarrow x = \pm 4$ $\Rightarrow y = \frac{3 (\pm 4)}{4} = \pm 3$ $y = \frac{4 x}{3} :$ $(i) \Rightarrow (x^{2} + x (\frac{4 x}{3}) + {(\frac{4 x}{3})}^{2}) \sqrt{x^{2} + {(\frac{4 x}{3})}^{2}} = 185$ $\frac{9 x^{2} + 12 x^{2} + 16 x^{2}}{9} \sqrt{\frac{9 x^{2} + 16 x^{2}}{9}} = 185$ $\frac{37 x^{2}}{9} \cdot \frac{5 ∣ x ∣}{3} = 185$ $∣ x ∣^{3} = \frac{185 \times 27}{37 \times 5} \Rightarrow∣ x ∣= 3 \Rightarrow x = \pm 3$ $\Rightarrow y = \frac{4 (\pm 3)}{3} = \pm 4$ $(x, y) = {(4, 3), (3, 4), (- 4, - 3), (- 3, - 4)}$
	Answered by mr W last updated on 02/Dec/22

	$u = \sqrt{x^{2} + y^{2}}$ $v = x y$ $(u^{2} + v) u = 185 . . . (i)$ $(u^{2} - v) u = 65 . . . (i i)$ $2 u^{3} = 185 + 65 = 250$ $\Rightarrow u = 5 = \sqrt{x^{2} + y^{2}}$ $\Rightarrow v = \frac{185}{5} - 5^{2} = 12 = x y$ $x^{2} + y^{2} = 25 \Rightarrow {(x + y)}^{2} = 25 + 2 \times 12 = 49$ $\Rightarrow x + y = \pm 7$ $\Rightarrow (x, y) = (3, 4) o r (4, 3) o r (- 3, - 4) o r (- 4, - 3)$
	Commented by Acem last updated on 02/Dec/22

	$N e a t S o l u t i o n!$
	Answered by Rasheed.Sindhi last updated on 02/Dec/22
	$(x^2 +xy+y^2 )(√(x^2 +y^2 )) =185....(i) (x^2 −xy+y^2 )(√(x^2 +y^2 )) =65....(ii) [(√(x^2 +y^2 )) ≠0] (i)×13 & (ii)×37 13(x^2 +xy+y^2 )(√(x^2 +y^2 )) =2405...(iii) 37(x^2 −xy+y^2 )(√(x^2 +y^2 )) =2405...(iv) (iii)&(iv): 13(x^2 +xy+y^2 )=37(x^2 −xy+y^2 ) 12x^2 −25xy+12y^2 =0 ((12x^2 )/(25xy))−1+((12y^2 )/(25xy))=0 ((12)/(25))((x/y)+(y/x))=1 t+(1/t)=((25)/(12)) 12t^2 −25t+12=0 t=((25±(√(625−576)))/(24))=((25±7)/(24))=(4/3),(3/4) (x/y)=(4/3),(3/4) 3x=4y ∨ 4x=3y y= ((3x)/4),((4x)/3) y=((3x)/4): (i)⇒(x^2 +x(((3x)/4))+(((3x)/4))^2 )(√(x^2 +(((3x)/4))^2 )) =185 ((16x^2 +12x^2 +9x^2 )/(16))(√((16x^2 +9x^2 )/(16))) =185 ((37x^2 )/(16))∙((5∣x∣)/4)=185 ∣x∣^3 =64 x=±4⇒y=((3(±4))/4)=±3 y= ((4x)/3): (i)⇒(x^2 +x(((4x)/3))+(((4x)/3))^2 )(√(x^2 +(((4x)/3))^2 )) =185 ((9x^2 +12x^2 +16x^2 )/9)(√((9x^2 +16x^2 )/9)) =185 ((37x^2 )/9)∙((5∣x∣)/3)=185 ∣x∣^3 =27⇒x=±3 ⇒y=((4(±3))/3)=±4 (x,y)={(3,4),(4,3),(−3,−4),(−4,−3)}$
	$(x^{2} + x y + y^{2}) \sqrt{x^{2} + y^{2}} = 185 . . . . (i)$ $(x^{2} - x y + y^{2}) \sqrt{x^{2} + y^{2}} = 65 . . . . (i i)$ $[\sqrt{x^{2} + y^{2}} \neq 0]$ $(i) \times 13 & (i i) \times 37$ $13 (x^{2} + x y + y^{2}) \sqrt{x^{2} + y^{2}} = 2405 . . . (i i i)$ $37 (x^{2} - x y + y^{2}) \sqrt{x^{2} + y^{2}} = 2405 . . . (i v)$ $(i i i) & (i v) :$ $13 (x^{2} + x y + y^{2}) = 37 (x^{2} - x y + y^{2})$ $12 x^{2} - 25 x y + 12 y^{2} = 0$ $\frac{12 x^{2}}{25 x y} - 1 + \frac{12 y^{2}}{25 x y} = 0$ $\frac{12}{25} (\frac{x}{y} + \frac{y}{x}) = 1$ $t + \frac{1}{t} = \frac{25}{12}$ $12 t^{2} - 25 t + 12 = 0$ $t = \frac{25 \pm \sqrt{625 - 576}}{24} = \frac{25 \pm 7}{24} = \frac{4}{3}, \frac{3}{4}$ $\frac{x}{y} = \frac{4}{3}, \frac{3}{4}$ $3 x = 4 y \lor 4 x = 3 y$ $y = \frac{3 x}{4}, \frac{4 x}{3}$ $y = \frac{3 x}{4} :$ $(i) \Rightarrow (x^{2} + x (\frac{3 x}{4}) + {(\frac{3 x}{4})}^{2}) \sqrt{x^{2} + {(\frac{3 x}{4})}^{2}} = 185$ $\frac{16 x^{2} + 12 x^{2} + 9 x^{2}}{16} \sqrt{\frac{16 x^{2} + 9 x^{2}}{16}} = 185$ $\frac{37 x^{2}}{16} \cdot \frac{5 ∣ x ∣}{4} = 185$ $∣ x ∣^{3} = 64$ $x = \pm 4 \Rightarrow y = \frac{3 (\pm 4)}{4} = \pm 3$ $y = \frac{4 x}{3} :$ $(i) \Rightarrow (x^{2} + x (\frac{4 x}{3}) + {(\frac{4 x}{3})}^{2}) \sqrt{x^{2} + {(\frac{4 x}{3})}^{2}} = 185$ $\frac{9 x^{2} + 12 x^{2} + 16 x^{2}}{9} \sqrt{\frac{9 x^{2} + 16 x^{2}}{9}} = 185$ $\frac{37 x^{2}}{9} \cdot \frac{5 ∣ x ∣}{3} = 185$ $∣ x ∣^{3} = 27 \Rightarrow x = \pm 3$ $\Rightarrow y = \frac{4 (\pm 3)}{3} = \pm 4$ $(x, y) = {(3, 4), (4, 3), (- 3, - 4), (- 4, - 3)}$
	Commented by Acem last updated on 02/Dec/22
	$Mr. Rasheed thank you for your efforts! Anyway, if you didn′t assume new variable, you can solve it directly without multiply the formulas by 13, 37 As you see if we sum i, ii we will save a lot of work (x^2 +y^2 ) (√(x^2 +y^2 ))= 125 ⇒ x^2 +y^2 = 25 ... iii iii ∧ ii: 25−xy= 13 ⇒ xy= 12 ... iv Now, (x+y)^2 = x^2 +y^2 + 2xy= 49 ⇒ x+y= ±7 ... v iv ∧ v: ∴ { ((x+y= 7)),(( xy= 12)) :} or { ((x+y= −7)),(( xy= 12)) :} ⇒ S={(4,3), (3,4), (−4, −3), (−3,−4)}$
	$M r . R a s h e e d t h a n k y o u f o r y o u r e f f o r t s!$ $A n y w a y, i f y o u {d i d n}^{'} t a s s u m e n e w v a r i a b l e, y o u$ $c a n s o l v e i t d i r e c t l y w i t h o u t m u l t i p l y t h e$ $f o r m u l a s b y 13, 37$ $A s y o u s e e i f w e s u m i, i i w e w i l l s a v e a l o t o f w o r k$ $(x^{2} + y^{2}) \sqrt{x^{2} + y^{2}} = 125 \Rightarrow x^{2} + y^{2} = 25 . . . i i i$ $i i i \land i i : 25 - x y = 13 \Rightarrow x y = 12 . . . i v$ $N o w, {(x + y)}^{2} = x^{2} + y^{2} + 2 x y = 49$ $\Rightarrow x + y = \pm 7 . . . v$ $i v \land v :$ $∴ {\begin{cases} x + y = 7 \\ x y = 12 \end{cases} o r {\begin{cases} x + y = - 7 \\ x y = 12 \end{cases}$ $\Rightarrow S = {(4, 3), (3, 4), (- 4, - 3), (- 3, - 4)}$
	Commented by Rasheed.Sindhi last updated on 02/Dec/22

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