Lagrangian formula. Can't find x and y

Question

$x+3y+2+λ(3x^2 +8xy +6y^2 - 18) =0$

$x+3y+2+3x^2 λ +8xyλ +6y^2 λ - 18λ =0$

$L'x =0$ $6xλ+8yλ +1=0$

$L'y =0$ $8xλ+12yλ+3=0$

$L'λ =0$ $3x^2 +8xy+6y^2 - 18=0$

I need to find x and y

Answer 1

We have
$$6\lambda x + 8\lambda y + 1 =0$$ $$8\lambda x + 12\lambda y + 3 = 0$$

By cross-multiplication method,

$$\color{blue}x = \frac{(24-12)\lambda}{(72-64)\lambda^2}=\color{blue}{\frac{3}{2\lambda}} \ , \color{blue}y = \frac{(8-18)}{(72-64)\lambda^2} = \color{blue}{\frac{-5}{4\lambda}}$$

Use $x$ and $y$ in terms of $\lambda$ in the third equation to get $\lambda$, then use the above relations to get $x$ and $y$.

Answer 2

Note one can not have $\lambda=0$. Take the $L_{x}=0$ equation and multiply by $3$ and subtract $L_{y}$ from it, to obtain

\begin{align*} 3(6x\lambda+8y\lambda+1)-(8x\lambda+12y\lambda+3)&=0\\ \left(10x+12y\right)\lambda&=0\\ x&=-\frac{6}{5}y\\ \end{align*}

Substitute this into the constraint equation $L_{\lambda}$ to get a quadratic equation in one variable, then backsubstitute.