System of equation with lambda How would I solve this system of equation? I guess Lagrange Multiplier is important when solving this.
$$\begin{align}&y^2=5\lambda\\
&2xy=3\lambda\\
&3x+4y=11
\end{align}$$
I have tried to solve this for one hour now but some reason I still don't seem to get in the right path. 
Could someone please help me with this one?
EDIT: After shuffling all the numbers I get Then $ x= \frac{33}{49}$ and  $y=\frac{110}{49}$. 
 A: You can “eliminate $\lambda$: multiply the first equation by $3$ and the second equation by $5$; then subtract the second from the first, getting
$$
3y^2-10xy=0
$$
Thus either $y=0$ or $y=10x/3$. Can you finish?
A: You can try to divided the first two equation. Then $\frac{y}{2x}=\frac53$. Can you finish it?
A: We have
$$y^2=5\lambda,\; 2xy=3\lambda \implies3y^2=10xy \implies y(3y-10x)=0$$
then consider two cases


*

*$y=0$ not acceptable

*$y\neq 0 \implies y=\frac{10}3 x\implies 3x+4\frac{10}3 x=11 \implies x=\cdots$
A: Put  $ 3x + 4y =  11 $ in terms of $ x $ and substitute into $ 2xy = 3L $. Then substitute the resulting equation into $ y^2  = 5L $ to arrive at
$$ 22y - 13y^2 = 0 $$
From there you can use the quadratic formula to find $y$.
Once you have y, substitute the value found in the equations and solve for $x$ and $L$
A: If $x=0$, we get by the third equation that $y=11/4$ and then the value of $\lambda$ follows by the first equation. Then we assume $x\neq 0$ and by the second equation $y=(3\lambda)/(2x)$. Using this, we get a system of two equations in $x$ and $\lambda$:
$9\lambda^2=(5\lambda)(4x^2)$ and $3x^2+6\lambda=11x$.
Now, again, if $\lambda=0$ we get the value of $x$. Then, again, we assume $\lambda\neq 0$, so that, by $9\lambda^2=(5\lambda)(4x^2)$, we get that $\lambda=(20/9)x^2$ and we can determine the value of $x$ using $3x^2+6\lambda=11x$.
