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Question Number 155106 by mathdanisur last updated on 25/Sep/21
$Solve for real numbers: { (((√(x+y)) - (√(x-y)) + (√(x^2 -y^2 )) = 5)),((2x + 3(√(x^2 -y^2 )) = 19)) :}$
$Solve for real numbers :$ ${\begin{cases} \sqrt{x + y} - \sqrt{x - y} + \sqrt{x^{2} - y^{2}} = 5 \\ 2 x + 3 \sqrt{x^{2} - y^{2}} = 19 \end{cases}$
Commented by benhamimed last updated on 25/Sep/21
$on pose a=(√(x+y)) b=(√(x−y)) ;tq a≥0 b≥0;x>y { ((a−b+ab=5)),((a^2 +b^2 +3ab=19)) :} { (((a−b)^2 =(5−ab)^2 ...(1))),(((a+b)^2 =19−ab...(2))) :} (2)−(1) →4ab=−25+19−ab+10ab−a^2 b^2 −(ab)^2 +5(ab)−6=0 Δ=25−4(−1)(−6)=1 ab=3 ∨ ab=2 si ab=3 { ((a−b=2)),(((a+b)^2 =16)) :} { ((a=3)),((b=1)) :}⇒ { ((x=5)),((y=4)) :} si ab=2 { ((a−b=3)),(((a+b)^2 =17)) :} ⇒ { ((a=(((√(17))+3)/2))),((b=(((√(17))−3)/2))) :} ⇒ { ((x=((13)/2))),((y=((3(√(17)))/2))) :} (x;y)={(5;4);(((13)/2);((3(√(17)))/2))}$
$o n p o s e a = \sqrt{x + y} b = \sqrt{x - y}; t q a ⩾ 0 b ⩾ 0; x > y$ ${\begin{cases} a - b + a b = 5 \\ a^{2} + b^{2} + 3 a b = 19 \end{cases}$ ${\begin{cases} {(a - b)}^{2} = {(5 - a b)}^{2} . . . (1) \\ {(a + b)}^{2} = 19 - a b . . . (2) \end{cases}$ $(2) - (1) \to 4 a b = - 25 + 19 - a b + 10 a b - a^{2} b^{2}$ $- {(a b)}^{2} + 5 (a b) - 6 = 0$ $Δ = 25 - 4 (- 1) (- 6) = 1$ $a b = 3 \lor a b = 2$ $s i a b = 3$ ${\begin{cases} a - b = 2 \\ {(a + b)}^{2} = 16 \end{cases} {\begin{cases} a = 3 \\ b = 1 \end{cases} \Rightarrow {\begin{cases} x = 5 \\ y = 4 \end{cases}$ $s i a b = 2$ ${\begin{cases} a - b = 3 \\ {(a + b)}^{2} = 17 \end{cases} \Rightarrow {\begin{cases} a = \frac{\sqrt{17} + 3}{2} \\ b = \frac{\sqrt{17} - 3}{2} \end{cases} \Rightarrow {\begin{cases} x = \frac{13}{2} \\ y = \frac{3 \sqrt{17}}{2} \end{cases}$ $(x; y) = {(5; 4); (\frac{13}{2}; \frac{3 \sqrt{17}}{2})}$
Commented by mathdanisur last updated on 25/Sep/21

$cool solution thankyou my dear$
Commented by Tawa11 last updated on 25/Sep/21

$Great sirs$
	Answered by MJS_new last updated on 25/Sep/21
	$x=((u^2 +v^2 )/2)∧y=((−u^2 +v^2 )/2)∧u≥0∧v≥0 ⇔ u=(√(x−y))∧v=(√(x+y)) { ((uv−u+v=5)),((u^2 +3uv+v^2 =19)) :} ⇒ { ((v=((u+5)/(u+1)))),(((u−1)(u+3)(u^2 +3u−2)=0)) :} ⇒ u=1∨u=−(3/2)+((√(17))/2) ⇒ v=3∨v=(3/2)+((√(17))/2) ⇒ x=5∧y=4∨x=((13)/2)∧y=((3(√(17)))/2)$
	$x = \frac{u^{2} + v^{2}}{2} \land y = \frac{- u^{2} + v^{2}}{2} \land u ⩾ 0 \land v ⩾ 0$ $\Leftrightarrow$ $u = \sqrt{x - y} \land v = \sqrt{x + y}$ ${\begin{cases} u v - u + v = 5 \\ u^{2} + 3 u v + v^{2} = 19 \end{cases}$ $\Rightarrow$ ${\begin{cases} v = \frac{u + 5}{u + 1} \\ (u - 1) (u + 3) (u^{2} + 3 u - 2) = 0 \end{cases}$ $\Rightarrow$ $u = 1 \lor u = - \frac{3}{2} + \frac{\sqrt{17}}{2}$ $\Rightarrow$ $v = 3 \lor v = \frac{3}{2} + \frac{\sqrt{17}}{2}$ $\Rightarrow$ $x = 5 \land y = 4 \lor x = \frac{13}{2} \land y = \frac{3 \sqrt{17}}{2}$
	Commented by mathdanisur last updated on 25/Sep/21

	$perfect solution thankyou my dear$
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