Samir's system of equations

$x^2 +y\sqrt{xy}=2013$ multiply with $\sqrt{y}$ become

$x^2\sqrt{y} +y^2\sqrt{x}=2013\sqrt{y}$ ..........(1)

$y^2 +x\sqrt{xy}=671$ multiply with $\sqrt{x}$ become

$y^2\sqrt{x} +x^2\sqrt{y}=671\sqrt{x}$ ..........(2)

Subtracting (1) and (2), we find

$2013\sqrt{y}-671\sqrt{x}=0$ , so

$x=9y$ , substitute to (2), we get

$28y^2=671$

$y^2=\frac {671}{28}$ , so $a+b=671+28=\underline{699}$

$x^2+y\sqrt{xy}=2013\Rightarrow {\sqrt{x}}(x\sqrt{x}+y\sqrt{y})=2013 - (1)$

$y^2+x\sqrt{xy}=671\Rightarrow {\sqrt{y}}(x\sqrt{x}+y\sqrt{y})=671 - (2)$

$\frac{(1)}{(2)}:\frac{\sqrt{x}}{\sqrt{y}}=3\Rightarrow \sqrt{x}=3\sqrt{y}-(3)$

Sub. (3) into (1)

$3\sqrt{y}[9y(3\sqrt{y})+y\sqrt{y}]=2013$

$28y^2=671\Rightarrow y^2=\frac{671}{28}$

$\therefore a + b = 699$

By factoring out $x^\frac{1}{2}$ on the top and $y^\frac{1}{2}$ you get $x^\frac{1}{2}(x^\frac{3}{2}+y^\frac{3}{2})=2013$ $y^\frac{1}{2}(x^\frac{3}{2}+y^\frac{3}{2})=671$ Dividing the second equation from the first gives $\frac{\sqrt{x}}{\sqrt{y}}=3$ or $x=9y$ , substituting back in to the first equation gives you $y^2=\frac{671}{28}$ , so $a+b=699$

Factor a $\sqrt{x}$ from the top equation and a $\sqrt{y}$ from the bottom: $(x\sqrt{x} + y\sqrt{y})\sqrt{x} = 2013 \\ (x\sqrt{x} + y\sqrt{y})\sqrt{y} = 671$ so we find that $\sqrt{x} = \frac{2013}{x\sqrt{x} + y\sqrt{y}} \\ \sqrt{y} = \frac{671}{x\sqrt{x} + y\sqrt{y}} = 3\sqrt{x}.$ So, when we square the final equation $\sqrt{y} = 3\sqrt{x}$ , we find that $y = 9x$ . Substituting the value of $x$ into the first equation, we find that $\begin{aligned} 81y^2 + y\sqrt{9y^2} &= 2013 \\ 27y^2 + y^2 &= 671 \\ y^2 &= \frac{671}{28}, \end{aligned}$ and $a + b = \boxed{699}$ .

$\frac{x^2+y\sqrt{xy}}{y^2+x\sqrt{xy}}=\frac{2013}{671}=3$ , so $\frac{x^2+y\sqrt{xy}}{y^2+x\sqrt{xy}} \cdot \frac{y^2-x\sqrt{xy}}{y^2-x\sqrt{xy}}=\frac{(y^3-x^3)\sqrt{3}}{y(y^3-x^3)}=\frac{\sqrt{xy}}{y}=\frac{2013}{671}=3$ . This is equivalent to $x=9y$ . Plugging back in to the second equation, we obtain $y^2+9y\sqrt{9y^2}=671$ , which yields $y^2=\frac{671}{28}$ .

The given equations can be transformed as $\sqrt{x}( x\sqrt{x} + y\sqrt{y}) = 2013 \\ \sqrt{y} ( y\sqrt{y} + x\sqrt{x}) = 671$ Dividing the first equation by the second yields $x = 9y$ . Plugging this value in any of the given equation and solving for $y^2$ gives us $y^2 = \frac{28}{671}$ $$ So $\boxed { a+b=699}$

GIven: $x^2 + y\sqrt{xy}=2013$ --eqn. (i)
$y^2 + x\sqrt{xy}=671$ --eqn. (ii)

Taking $\sqrt{x}$ common from eqn (i) and $\sqrt{y}$ from eqn (ii), we get - $\sqrt{x}(x\sqrt{x} + y\sqrt{y})=2013$ --eqn. (iii)
$\sqrt{y}(y\sqrt{y} + x\sqrt{x}) = 671$ --eqn. (iv)

Now we divide eqn(iii) by eqn(iv), we are left with - $\sqrt{\frac{x}{y}} = 3 \Rightarrow \sqrt{x} = 3\sqrt{y} \Rightarrow x = 9y$

Now substituting the values of $\sqrt{x}$ and $x$ in eqn(ii) , we get -

$y^2 + 27y^2 = 671 \Rightarrow y^2 = \frac{671}{28}$ Since 671 and 28 are coprime to each other, $a=671, b=28 \Rightarrow a+b=699$

Let $x=ty$ for some positive $t$ . Our equations are: $t^2y^2 + y\sqrt{ty^2}=y^2(t^2+\sqrt{t})=2013$ and

$y^2+ty\sqrt{ty^2}=y^2(1+t\sqrt{t})=671$

$\Rightarrow \frac{671 \sqrt{t}}{\sqrt{t}+t^2}=\frac{2013}{\sqrt{t}+t^2}$ $\sqrt{t}=3 \Rightarrow t=9$

$y^2=\frac{671}{1+9 \cdot 3}=\frac{671}{28}$

Hence, since $gcd(671,28)=1$ , $a+b=\boxed{699}$ .

Therefore 671+28=699

x² + ysqrt(xy) = 2013 (I) y² + xsqrt(xy) = 671 (II)

Multiply (II) by 3 and then get:

x² + ysqrt(xy) = 3y² + 3xsqrt(xy)

x²-3y² = (3x-y)*sqrt(xy)

sqrt(xy) = (x²-3y²)/(3x-y)

Squaring both sides:

xy = (x²-3y²)²/(3x-y)²

xy(3x-y)² = (x²-3y²)², which gives us x = 9y.

Make the substitution and then find y² = 671/28. Hence, a + b = 699.

Eq 1 - 3* Eq 2 = 0 will be an homogeneous equation, y<>0, we can divide by y^2. Let x/y=t^2; t > 0. We obtain t=3. On the other hand, if x = y * t^2, from Eq 1 we have y^2=2013 / (t^4+t). So y^2 = 671 / 28, a, b coprimes ... a+b=699 :)

Realize that $\frac{x^{2}+y\sqrt{xy}}{y^{2}+x\sqrt{xy}}=3$

We multiply the numerator and the denominator by $y^2-x\sqrt{xy}$ .

We get: $\frac{(x^2+y\sqrt{xy})(y^2-x\sqrt{xy})}{(y^2+x\sqrt{xy})(y^2-x\sqrt{xy})}=3$

We simplify the expression above.

$\frac{x^2y^2+(y^3-x^3) \sqrt{xy} - x^2y^2}{y(y^3-x^3)}=3$

We can now get a much more simple equation. $\frac{\sqrt{xy}}{y}=3 \rightarrow x=9y$

Subtitute the value of $x$ to the second equation.

$y^2 + x\sqrt{xy} = 671$

Subtituting, $y^2 + 9y \cdot \sqrt{9y\cdot y} =671$ .

It is simple to work out the value of $y$ .

$y^2 + 27 y^2 = 671$

$y^2=\frac{671}{28} \rightarrow a+b=\boxed{699}$

First note that both the equations can be written as; $x \sqrt{x} \sqrt(x)+ \sqrt(x)y \sqrt(y)=2013 \Rightarrow \sqrt(x)(x \sqrt(x)+y \sqrt(y))=2013$

$y \sqrt(y) \sqrt(y)+ x \sqrt(x) \sqrt(y)=671 \Rightarrow \sqrt(y)(y \sqrt(y)+x \sqrt(x))=671$

As $y \sqrt(y)+x \sqrt(x)>0$ we can divide both euations to get;

$\frac{\sqrt(x)}{\sqrt(y)=3} \Rightarrow \frac{x}{y}=9 \Rightarrow x=9y$

If so we can then get; $(9y)^2+3y^2=2013 \Rightarrow 84y^2=2013 \Rightarrow y^2=\frac{2013}{84}=\frac{671}{28}$

So we get $671+28=699$

We have $\sqrt{x}(x\sqrt{x}+y\sqrt{y})=2013$ and $\sqrt{y}(y\sqrt{y}+x\sqrt{x})=671$ . So we will get $\sqrt{x}=3\sqrt{y}$ or equivalent with $x=9y$ . Substitute it to the first equation, then we get $81y^2+3y^2=2013$ , or equivalently $y^2=\dfrac{2013}{84}= \dfrac{671}{28}$ . So $a+b=671+28=699$

We can note that $\sqrt{xy}$ is the geometric mean of $x$ and $y$ , which implies that $(y, \sqrt{xy}, x)$ is a geometric progression.

Now since we are to find the value of $y^{2}$ , we can express $x$ and $\sqrt{xy}$ in terms of $y$ and $r$ , the common ratio. Since $\sqrt{xy}$ is the geometric mean it must be expressed as: $\sqrt{xy} = yr$ and $x$ as the third term so: $x = yr^{2}$

Substituting it to the first and second equation, it will yield:

$y^{2}r^{4}+ y^{2}r = 2013$

$y^{2} + y^{2}r^{3} = 671$

Which can be factored,

$y^{2}r(1 + r^{3}) = 2013$

$y^{2}(1 + r^{3}) = 671$

Now dividing the first equation to the second equation will yield the common ratio $r = 3$

And finally substituting to the first equation we can finally find the value of $y^{2}$

$y^{2} (3 + 3^{4}) = 2013$

$y^{2} = \frac{2013}{84}$

$y^{2} = \frac{671}{28}$

Thus, $671 + 28 = 699$

Remark that $2013=3×671$ .

(First equation)- 3×(Second equation) = 0 then.

Divide both sides by $\sqrt{xy}$ and set $A=x/y$ .

Then $A(\sqrt{A}-3)+1-\frac{3}{\sqrt{A}} = 0$

This equation has only $A=9$ as a solution.

This implies $x=9y$

Plug it in back in the equations above to get $y^{2}=671/28$

Hence the result : 699

Let $x = my$ for some real number $m$ .

Then we have from the first equation ,

$m^2y^2 + y^2\sqrt{m} = 2013$

$\Rightarrow y^2\sqrt{m}(m\sqrt{m} + 1) = 2013 ......(1)$

Similarly from second equation we have

$y^2(m\sqrt{m} +1) = 671 .........(2)$

Dividing $(1)$ by $(2)$

$\sqrt{m} = 3$

$\Rightarrow m = 9$

Substituting $m$ in $(2)$ , we have

$28y^2 = 671$

$\Rightarrow y^2 = \frac{671}{28}$

$a + b = 671 +28 = 699$