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Solve-for-real-numbers-x-y-x-y-x-2-y-2-5-2x-3-x-2-y-2-19-

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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)) :}
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\sqrt{\mathrm{x}+\mathrm{y}}\:-\:\sqrt{\mathrm{x}-\mathrm{y}}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} -\mathrm{y}^{\mathrm{2}} }\:=\:\mathrm{5}}\\{\mathrm{2x}\:+\:\mathrm{3}\sqrt{\mathrm{x}^{\mathrm{2}} -\mathrm{y}^{\mathrm{2}} }\:=\:\mathrm{19}}\end{cases} \\ $$
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))}
$${on}\:{pose}\:{a}=\sqrt{{x}+{y}}\:\:\:{b}=\sqrt{{x}−{y}}\:\:;{tq}\:{a}\geqslant\mathrm{0}\:{b}\geqslant\mathrm{0};{x}>{y} \\ $$$$\begin{cases}{{a}−{b}+{ab}=\mathrm{5}}\\{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}{ab}=\mathrm{19}}\end{cases} \\ $$$$\begin{cases}{\left({a}−{b}\right)^{\mathrm{2}} =\left(\mathrm{5}−{ab}\right)^{\mathrm{2}} …\left(\mathrm{1}\right)}\\{\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{19}−{ab}…\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{2}\right)−\left(\mathrm{1}\right)\:\:\rightarrow\mathrm{4}{ab}=−\mathrm{25}+\mathrm{19}−{ab}+\mathrm{10}{ab}−{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$−\left({ab}\right)^{\mathrm{2}} +\mathrm{5}\left({ab}\right)−\mathrm{6}=\mathrm{0} \\ $$$$\Delta=\mathrm{25}−\mathrm{4}\left(−\mathrm{1}\right)\left(−\mathrm{6}\right)=\mathrm{1} \\ $$$${ab}=\mathrm{3}\:\:\:\vee\:\:\:{ab}=\mathrm{2} \\ $$$${si}\:{ab}=\mathrm{3} \\ $$$$\begin{cases}{{a}−{b}=\mathrm{2}}\\{\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{16}}\end{cases}\:\:\:\begin{cases}{{a}=\mathrm{3}}\\{{b}=\mathrm{1}}\end{cases}\Rightarrow\begin{cases}{{x}=\mathrm{5}}\\{{y}=\mathrm{4}}\end{cases} \\ $$$${si}\:{ab}=\mathrm{2} \\ $$$$\begin{cases}{{a}−{b}=\mathrm{3}}\\{\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{17}}\end{cases}\:\Rightarrow\begin{cases}{{a}=\frac{\sqrt{\mathrm{17}}+\mathrm{3}}{\mathrm{2}}}\\{{b}=\frac{\sqrt{\mathrm{17}}−\mathrm{3}}{\mathrm{2}}}\end{cases}\:\:\Rightarrow\begin{cases}{{x}=\frac{\mathrm{13}}{\mathrm{2}}}\\{{y}=\frac{\mathrm{3}\sqrt{\mathrm{17}}}{\mathrm{2}}}\end{cases} \\ $$$$\left({x};{y}\right)=\left\{\left(\mathrm{5};\mathrm{4}\right);\left(\frac{\mathrm{13}}{\mathrm{2}};\frac{\mathrm{3}\sqrt{\mathrm{17}}}{\mathrm{2}}\right)\right\} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 25/Sep/21
cool solution thankyou my dear
$$\mathrm{cool}\:\mathrm{solution}\:\mathrm{thankyou}\:\mathrm{my}\:\mathrm{dear} \\ $$
Commented by Tawa11 last updated on 25/Sep/21
Great sirs
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
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)
$${x}=\frac{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }{\mathrm{2}}\wedge{y}=\frac{−{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }{\mathrm{2}}\wedge{u}\geqslant\mathrm{0}\wedge{v}\geqslant\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${u}=\sqrt{{x}−{y}}\wedge{v}=\sqrt{{x}+{y}} \\ $$$$\begin{cases}{{uv}−{u}+{v}=\mathrm{5}}\\{{u}^{\mathrm{2}} +\mathrm{3}{uv}+{v}^{\mathrm{2}} =\mathrm{19}}\end{cases} \\ $$$$\Rightarrow \\ $$$$\begin{cases}{{v}=\frac{{u}+\mathrm{5}}{{u}+\mathrm{1}}}\\{\left({u}−\mathrm{1}\right)\left({u}+\mathrm{3}\right)\left({u}^{\mathrm{2}} +\mathrm{3}{u}−\mathrm{2}\right)=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow \\ $$$${u}=\mathrm{1}\vee{u}=−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${v}=\mathrm{3}\vee{v}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{5}\wedge{y}=\mathrm{4}\vee{x}=\frac{\mathrm{13}}{\mathrm{2}}\wedge{y}=\frac{\mathrm{3}\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 25/Sep/21
perfect solution thankyou my dear
$$\mathrm{perfect}\:\mathrm{solution}\:\mathrm{thankyou}\:\mathrm{my}\:\mathrm{dear} \\ $$

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