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.

EDIT: After shuffling all the numbers I get Then $$x= \frac{33}{49}$$ and $$y=\frac{110}{49}$$.

• Following the given hint we have y=x10/3 for sure. From here we can easily find x and y. Therefore your solution seems correct.
– user
Sep 26, 2019 at 13:49

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?

You can try to divided the first two equation. Then $$\frac{y}{2x}=\frac53$$. Can you finish it?

• As a final solution I get $x= \frac{33}{49}$ and $y=\frac{110}{49}$. Do you think this is correct? Sep 26, 2019 at 13:45
• @MiMaKo No, you're forgetting a solution. Sep 26, 2019 at 22:12
• @egreg Okay, that's cryptic. Do you mean lambda is the correct answer? If so, then it is $\lambda=\frac{2420}{2401}$ Sep 26, 2019 at 22:27
• @MiMaKo One cannot divide by zero, which is the problem with this answer. If you do it correctly, you get either $y=0$ (hence $x=11/3$ and $\lambda=0$) or $y=10x/3$ (so $x=33/49$, $y=110/49$ and $\lambda=2420/2401$). Sep 26, 2019 at 22:38

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 $$L$$

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
• $$y\neq 0 \implies y=\frac{10}3 x\implies 3x+4\frac{10}3 x=11 \implies x=\cdots$$

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 $$3x^2+6\lambda=11x$$.