Can This Unique Pair of Values Solve the Equation?

[anemone]

Problem:
Solve the system:
$\displaystyle \sqrt{3x}\left(1+\frac{1}{x+y} \right)=2$

$\displaystyle \sqrt{7y}\left(1-\frac{1}{x+y} \right)=4\sqrt{2}$

 

[Albert1]

y=6x
x approaches 1.03
y approaches 6.18
in fact x is the solution of quadratic equation :
$147x^2-154x+3=0$
we delete the smaller solution

 

[anemone]

y=6x
x approaches 1.03
y approaches 6.18
in fact x is the solution of quadratic equation :
$147x^2-154x+3=0$
we delete the smaller solution

I'm sorry Albert, I don't follow your reasoning at all...

 

[Fernando Revilla]

We can express $$\left\{\begin{matrix}
\sqrt{3x}\left(1+\dfrac{1}{x+y}\right)=2\\ \sqrt{7y}\left(1-\dfrac{1}{x+y}\right)
=4\sqrt{2} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 1+\dfrac{1}{x+y}=\dfrac{2}{\sqrt{3x}}\quad (1)\\ 1-\dfrac{1}{x+y}=\dfrac{4\sqrt{2}}{\sqrt{7y}}\quad (2)\end{matrix}\right.$$ Using the transformations $(1)+(2)$ and $(1)-(2)$: $$\left\{\begin{matrix} 1=\frac{1}{\sqrt{3x}}+\frac{2\sqrt{2}}{\sqrt{7y}} \quad (3)\\ \frac{1}{x+y}=\frac{1}{\sqrt{3x}}-\frac{2\sqrt{2}}{\sqrt{7y}}\quad (4) \end{matrix}\right.$$ Multiplying $(3)$ by $(4)$: $$\frac{1}{x+y}=\frac{1}{3x}-\frac{8}{7y}=\frac{7y-24x}{21xy}\Leftrightarrow\ldots \Leftrightarrow 7y^{2}-38xy-24x^{2}=0\quad (5)$$ Solving the quadratic equation $(5)$ on $y$: $$y=\dfrac{38x\pm\sqrt{2116x^2}}{14}=\ldots=\{6x,-(4/7)x\}$$ Hence, we get $y=6x$ or $y=-\frac{4}{7}x$. Could you continue?

 

[Fernando Revilla]

y=6x
x approaches 1.03
y approaches 6.18
in fact x is the solution of quadratic equation :
$147x^2-154x+3=0$
we delete the smaller solution

Why do you start with $y=6x$?

 

[Albert1]

Let x+y=A
we get :
$\sqrt{\dfrac{4}{3x}}=1+\dfrac{1}{A}$-----------(1)
$\sqrt{\dfrac{32}{7y}}=1-\dfrac{1}{A}$----------(2)
$(1)^2-(2)^2 :$
$\dfrac{4}{3x}-\dfrac{32}{7y}=\dfrac{4}{x+y}$
$\dfrac{1}{(x+y)}=\dfrac{(7y-24x)}{(21xy)}$
$24x^2+38xy-7y^2$=0
(4x+7y)(6x-y)=0
y=$\dfrac{-4x}{7}$ or y=6x
because x>0 and y>0
so y=$\dfrac{-4x}{7}$ does not fit
so we start with y=6x

 

[anemone]

We can express $$\left\{\begin{matrix}
\sqrt{3x}\left(1+\dfrac{1}{x+y}\right)=2\\ \sqrt{7y}\left(1-\dfrac{1}{x+y}\right)
=4\sqrt{2} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 1+\dfrac{1}{x+y}=\dfrac{2}{\sqrt{3x}}\quad (1)\\ 1-\dfrac{1}{x+y}=\dfrac{4\sqrt{2}}{\sqrt{7y}}\quad (2)\end{matrix}\right.$$ Using the transformations $(1)+(2)$ and $(1)-(2)$: $$\left\{\begin{matrix} 1=\frac{1}{\sqrt{3x}}+\frac{2\sqrt{2}}{\sqrt{7y}} \quad (3)\\ \frac{1}{x+y}=\frac{1}{\sqrt{3x}}-\frac{2\sqrt{2}}{\sqrt{7y}}\quad (4) \end{matrix}\right.$$ Multiplying $(3)$ by $(4)$: $$\frac{1}{x+y}=\frac{1}{3x}-\frac{8}{7y}=\frac{7y-24x}{21xy}\Leftrightarrow\ldots \Leftrightarrow 7y^{2}-38xy-24x^{2}=0\quad (5)$$ Solving the quadratic equation $(5)$ on $y$: $$y=\dfrac{38x\pm\sqrt{2116x^2}}{14}=\ldots=\{6x,-(4/7)x\}$$ Hence, we get $y=6x$ or $y=-\frac{4}{7}x$. Could you continue?

That's great, Fernando!

And yes, I could continue from there...

From $7y^2-38xy-24x^2=0$, we factorize it (like what Albert did, to get $(y-6x)(7y+4x)=0$ and we find $y=6x$ or $y=-\frac{4}{7}$.

Since $x, y>0$, we're left with only $y=6x$ to consider with.

If we substitute $y=6x$ back into the equation $\displaystyle 1+\frac{1}{x+y}=\frac{2}{\sqrt{3x}}$and letting $k=\sqrt{x}$, we obtain:

$7\sqrt{3}k^2-14k+\sqrt{3}=0$

Now, use the quadratic formula to solve for k yields $\displaystyle k=\frac{7\pm 2\sqrt{7}}{7\sqrt{3}}$, thus, $\displaystyle x=(\frac{7\pm 2\sqrt{7}}{7\sqrt{3}})^2$ and $\displaystyle y=6(\frac{7\pm 2\sqrt{7}}{7\sqrt{3}})^2$.

 

[Albert1]

according to your post , you have two solutions,but one of the combinations
will make $1-\dfrac{1}{x+y}<0$ ,this combination
should also be deleted

 

[anemone]

according to your post , you have two solutions,but one of the combinations
will make 1-1/(x+y) <0 ,this combination
should also be deleted

You're right, Albert! I didn't check and verify the answer. Thanks for letting me know about this.

We have only one pair of value that satisfies the given equations...and they're
$\displaystyle x=(\frac{7+ 2\sqrt{7}}{7\sqrt{3}})^2$ and $\displaystyle y=6(\frac{7+ 2\sqrt{7}}{7\sqrt{3}})^2$.