((BobHans)/(βo♭)) (1) { (((x+y)(x^2 −y^2 ) = 9)),(((x−y)(x^2 +y^2 ) = 5)) :} find the solution (2) x (dy/dx) = x^2 +y^2 when x=1 give y = 2


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Question Number 108295 by bobhans last updated on 16/Aug/20
$((BobHans)/(βo♭)) (1) { (((x+y)(x^2 −y^2 ) = 9)),(((x−y)(x^2 +y^2 ) = 5)) :} find the solution (2) x (dy/dx) = x^2 +y^2 when x=1 give y = 2$
$\frac{B ob H ans}{β o ♭}$ $(1) {\begin{cases} (x + y) (x^{2} - y^{2}) = 9 \\ (x - y) (x^{2} + y^{2}) = 5 \end{cases}$ $find the solution$ $(2) x \frac{dy}{dx} = x^{2} + y^{2} when x = 1 give y = 2$
	Answered by john santu last updated on 16/Aug/20
	$((⇈ JS ⇈)/♥) ⇒ { (((x+y)^2 (x−y) = 9)),((x−y = (5/(x^2 +y^2 )))) :} ⇒5(x+y)^2 = 9x^2 +9y^2 ⇒5x^2 +10xy+5y^2 = 9x^2 +9y^2 ⇒4x^2 −10xy+4y^2 =0 ⇒2x^2 −5xy+2y^2 =0 ⇒(2x−y)(x−2y)=0 { ((y=2x⇒3x.(−3x^2 )=9→ { ((x=−1)),((y=−2)) :})),((x=2y⇒3y.(3y^2 )=9→ { ((y=1)),((x=2)) :})) :} then solution set is {(−1,−2),(2,1)}$
	$\frac{⇈ J S ⇈}{♡}$ $\Rightarrow {\begin{cases} {(x + y)}^{2} (x - y) = 9 \\ x - y = \frac{5}{x^{2} + y^{2}} \end{cases}$ $\Rightarrow 5 {(x + y)}^{2} = 9 x^{2} + 9 y^{2}$ $\Rightarrow 5 x^{2} + 10 x y + 5 y^{2} = 9 x^{2} + 9 y^{2}$ $\Rightarrow 4 x^{2} - 10 x y + 4 y^{2} = 0$ $\Rightarrow 2 x^{2} - 5 x y + 2 y^{2} = 0$ $\Rightarrow (2 x - y) (x - 2 y) = 0$ ${\begin{cases} y = 2 x \Rightarrow 3 x . (- 3 x^{2}) = 9 \to {\begin{cases} x = - 1 \\ y = - 2 \end{cases} \\ x = 2 y \Rightarrow 3 y . (3 y^{2}) = 9 \to {\begin{cases} y = 1 \\ x = 2 \end{cases} \end{cases}$ $t h e n s o l u t i o n s e t i s {(- 1, - 2), (2, 1)}$
	Commented by bemath last updated on 16/Aug/20

	$c o o l l$
	Commented by bobhans last updated on 16/Aug/20

	$good$
	Answered by 1549442205PVT last updated on 16/Aug/20
	${ (((x+y)(x^2 −y^2 ) = 9)),(((x−y)(x^2 +y^2 ) = 5)) :} (1) If x−y=0 then LHS =0 ⇒the given system has no solutions,so x−y≠0 From (1)we have 5(x+y)(x^2 −y^2 ) =9(x−y)(x^2 +y^2 ) ⇔(x−y)[5(x+y)^2 −9(x^2 +y^2 )]=0 ⇔5(x+y)^2 −9(x^2 +y^2 )=0 ⇔4(x^2 +y^2 )−10xy=0.Put x=ky(k≠0) we get 4k^2 −10k+4=0⇔2k^2 −5k+2=0 ⇒k∈{2;1/2} i)For k=2⇒x=2y .Replce into first equation we get 3y.3y^2 =9⇔y^3 =1⇔y=1⇒x=2 we have solution (x,y)=(2,1) ii)For k=1/2⇔y=2x, replace into first equation we get −x.(−3x^2 )=9⇔x^3 =1⇔x=1⇒y=2 we get solution (x;y)=(1;2) Thus,given system of equations has two solutions (x;y)∈{(1,2);(2;1)}$
	${\begin{cases} (x + y) (x^{2} - y^{2}) = 9 \\ (x - y) (x^{2} + y^{2}) = 5 \end{cases} (1)$ $If x - y = 0 then LHS = 0 \Rightarrow the given$ $system has no solutions, so x - y \neq 0$ $From (1) we have 5 (x + y) (x^{2} - y^{2})$ $= 9 (x - y) (x^{2} + y^{2})$ $\Leftrightarrow (x - y) [5 {(x + y)}^{2} - 9 (x^{2} + y^{2})] = 0$ $\Leftrightarrow 5 {(x + y)}^{2} - 9 (x^{2} + y^{2}) = 0$ $\Leftrightarrow 4 (x^{2} + y^{2}) - 10 xy = 0 . Put x = ky (k \neq 0)$ $we get {4 k}^{2} - 10 k + 4 = 0 \Leftrightarrow {2 k}^{2} - 5 k + 2 = 0$ $\Rightarrow k \in {2; 1 / 2}$ $i) For k = 2 \Rightarrow x = 2 y . Replce into first$ $equation we get$ $3 y . {3 y}^{2} = 9 \Leftrightarrow y^{3} = 1 \Leftrightarrow y = 1 \Rightarrow x = 2 we have$ $solution (x, y) = (2, 1)$ $ii) For k = 1 / 2 \Leftrightarrow y = 2 x, replace into first equation$ $we get - x . (- {3 x}^{2}) = 9 \Leftrightarrow x^{3} = 1 \Leftrightarrow x = 1 \Rightarrow y = 2$ $we get solution (x; y) = (1; 2)$ $Thus, given system of equations has$ $two solutions (x; y) \in {(1, 2); (2; 1)}$
	Commented by bemath last updated on 16/Aug/20

	$i f (1, 2) \Rightarrow (1 + 2) (1 - 4) = - 9 \neq 9$ $w r o n g s i r$
	Commented by bemath last updated on 16/Aug/20

	$y e s s i r . t h a n k y o u$
	Commented by 1549442205PVT last updated on 16/Aug/20

	$Thank you . I mistaked . The cases k = \frac{1}{2}$ $should correct as : y = 2 x, so replace into$ $first eqn . \Rightarrow 3 x . (- {3 x}^{2}) = 9 \Leftrightarrow x^{3} = - 1$ $\Leftrightarrow x = - 1 \Rightarrow y = - 2 \Rightarrow (x, y) = (- 1, - 2)$
	Answered by Dwaipayan Shikari last updated on 16/Aug/20
	$(((x+y)(x^2 −y^2 ))/((x−y)(x^2 +y^2 )))=(9/5) (((x+y)^2 )/((x^2 +y^2 )))=(9/5) 1+((2xy)/(x^2 +y^2 ))=(9/5) ((2xy)/(x^2 +y^2 ))=(4/5) ((x^2 +y^2 )/(2xy))=(5/4) ((x^2 +y^2 +2xy)/(x^2 +y^2 −2xy))=(9/1) (((x+y)/(x−y)))^2 =9 ((x+y)/(x−y))=3 or ((x+y)/(x−y))=−3 (x/y)=2 or (x/y)=(1/2) , x=k_0 y=2k_(0 ) so 3k_0 .(−3k_0 ^2 )=9 ⇒k_0 =−1 x=2k y=k (x+y)(x^2 −y^2 )=9 3k.3k^2 =9 k^3 =1 k=1 { ((x=2)),((y=1)) :} or { ((x=−1)),((y=−2)) :}$
	$\frac{(x + y) (x^{2} - y^{2})}{(x - y) (x^{2} + y^{2})} = \frac{9}{5}$ $\frac{{(x + y)}^{2}}{(x^{2} + y^{2})} = \frac{9}{5}$ $1 + \frac{2 x y}{x^{2} + y^{2}} = \frac{9}{5}$ $\frac{2 x y}{x^{2} + y^{2}} = \frac{4}{5}$ $\frac{x^{2} + y^{2}}{2 x y} = \frac{5}{4}$ $\frac{x^{2} + y^{2} + 2 x y}{x^{2} + y^{2} - 2 x y} = \frac{9}{1}$ ${(\frac{x + y}{x - y})}^{2} = 9$ $\frac{x + y}{x - y} = 3 o r \frac{x + y}{x - y} = - 3$ $\frac{x}{y} = 2 o r \frac{x}{y} = \frac{1}{2}, x = k_{0} y = 2 k_{0} s o 3 k_{0} . (- 3 k_{0}^{2}) = 9 \Rightarrow k_{0} = - 1$ $x = 2 k$ $y = k$ $(x + y) (x^{2} - y^{2}) = 9$ $3 k . 3 k^{2} = 9$ $k^{3} = 1$ $k = 1$ ${\begin{cases} x = 2 \\ y = 1 \end{cases} o r {\begin{cases} x = - 1 \\ y = - 2 \end{cases}$
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