# System of two equations

Linear, quadratic, module, parametric equations

### System of two equations

$$\begin{cases} x + y = 5 \\ x\sqrt{y} - y\sqrt{x} = 6 \end{cases}$$
Guest

### Re: System of two equations

EDIT:
$$\begin{cases} x + y = 5 \\ x\sqrt{y} + y\sqrt{x} = 6 \end{cases}$$

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

antoniu200

Posts: 2
Joined: Mon Feb 25, 2019 12:07 pm
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### Re: System of two equations

Obvious first step: from x+ y= 5, y= 5- x. Replace "y" in the other equation by that. $$x\sqrt{y}+ y\sqrt{x}= x\sqrt{5- x}+ (5- x)\sqrt{x}= 6$$. 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

### Re: System of two equations

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

$$\begin{array}{|l} a^{2}+b^{2}=5 \\ a^{2}b+ab^{2}=6 \end{array}$$

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

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

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

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

$$\begin{array}{|l} \sqrt{x}=2 \\ \sqrt{y}=1 \end{array}$$ or $$\begin{array}{|l} \sqrt{x}=1 \\ \sqrt{y}=2\end{array}$$

(4;1) or (1;4)
Guest

### Re: System of two equations

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.

antoniu200

Posts: 2
Joined: Mon Feb 25, 2019 12:07 pm
Reputation: 0

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