System of two equations

[Guest]

[tex]\begin{cases} x + y = 5 \\ x\sqrt{y} - y\sqrt{x} = 6 \end{cases}[/tex]

[antoniu200]

EDIT:
[tex]\begin{cases} x + y = 5 \\ x\sqrt{y} + y\sqrt{x} = 6 \end{cases}[/tex]

Also, this is my post, sorry for the confusion, but somehow I got logged out when submitted this post.

[Guest]

Obvious first step: from x+ y= 5, y= 5- x. Replace "y" in the other equation by that. [tex]x\sqrt{y}+ y\sqrt{x}= x\sqrt{5- x}+ (5- x)\sqrt{x}= 6[/tex]. Now square both sides. That will have a single square root in it. Isolate that square root and square again. The result will be a fourth degree polynomial equation.

[Guest]

[tex]\begin{array}{|l} x + y = 5 \\ x\sqrt{y} + y\sqrt{x} = 6 \end{array}[/tex] [ 1 put ] [tex]\sqrt{x}[/tex]=a ; [tex]\sqrt{y}[/tex]=b

[tex]\begin{array}{|l} a^{2}+b^{2}=5 \\ a^{2}b+ab^{2}=6 \end{array}[/tex]

[tex]\begin{array}{|l} (a^{2}+2ab+b^{2})-2ab=5 \\ ab(a+b)=6 \end{array}[/tex] [ 2 put ] a+b=m ;ab=n

[tex]\begin{array}{|l} m^{2} -2n=5 [n=\frac{m^{2}-5}{2}]\\ mn=6 (A)\end{array}[/tex] [tex]\Rightarrow[/tex] [tex]\frac{m(m^{2}-5)}{2}[/tex]=6

[tex]m^{3}[/tex]-5m-12=0 [tex]\Leftrightarrow[/tex] [tex]m^{3}[/tex]+3[tex]m^{2}[/tex]+4m-3[tex]m^{2}[/tex]-9m-12=0
(m-3)([tex]m^{2}[/tex]+3m+4)=0 [tex]\Rightarrow[/tex] m=3 [of (A)] n=2

[tex]\begin{array}{|l} a+b=3 (B) \\ ab=2 \end{array}[/tex]
a(3-a)=2 [tex]\Leftrightarrow[/tex] [tex]a^{2}[/tex]-3a+2=0 ; D=9-8=1 ; [tex]a_{1,2 }[/tex]=[tex]\frac{3\pm1}{2}[/tex]
[tex]a_{1 }[/tex]=2 , [tex]a_{2 }[/tex]=1 [of (B)] [tex]b_{1 }[/tex]=1 , [tex]b_{2 }[/tex]=2 ;[ to 1 put ]

[tex]\begin{array}{|l} \sqrt{x}=2 \\ \sqrt{y}=1 \end{array}[/tex] or [tex]\begin{array}{|l} \sqrt{x}=1 \\ \sqrt{y}=2\end{array}[/tex]

(4;1) or (1;4)

[antoniu200]

Thanks to everyone who replied and helped! I was also shown a different method by a friend, but I don't have time right now to write it here: it basically solves any 3rd degree equation with a table.