If $a x_n + b y_n \rightarrow L$ and $c x_n + d y_n \rightarrow M$, then $x_n$ and $y_n$ are convergent. Prove or else give a counterexample: If $ad - bc \neq 0$ and if
\begin{align*}
    a x_n + b y_n \rightarrow L && \text{and} && c x_n + d y_n \rightarrow M
\end{align*}
as $n \rightarrow \infty$, then $x_n$ converges and $y_n$ converges.
I've been playing around with this one in a similar manner as I did here, but I cannot figure out how to make the $y_n$ term cancel like I did there, or find some other way to isolate the $x_n$ term.
It's easy enough to get to the expression
$-2 \epsilon < (a+c) x_n + (b+d) y_n - L - M < 2 \epsilon$ from the definition of a limit, but I'm not sure where to go from there. What I'd like to do is either subtract divide through by $(b+d) y_n$, but I don't think it's valid to include $y_n$ in the epsilon term while showing that $x_n$ meets the definition of convergence.
I've had just as little luck trying to show it from any of the sum/difference/product/quotient of the limit rules. We get something like $a c (x_n)^2 + a d x_n y_n + b c x_n y_n + b d (y_n)^2 \rightarrow LM$, and again there's the same problem of isolating the $x_n$ terms.
Any insights would be greatly appreciated.
 A: If $a=0$ then it's easy.
Let $a \neq 0$
$x_n+\frac{b}{a}y_n \to \frac{L}{a}$
$x_n+\frac{d}{c}y_n \to \frac{M}{c}$
If you substract you have the convergence of $y_n$ since $bc \neq ad$
Continue from here.
A: We can also use a little bit of linear algebra: Let $u_n=ax_n+by_n$ and $v_n=cx_n+dy_n$. Then we have that
$$\begin{bmatrix}
u_n\\
v_n
\end{bmatrix} = \begin{bmatrix}
a&b\\
c&d
\end{bmatrix} \begin{bmatrix}
x_n\\
y_n
\end{bmatrix}$$
And since $ad-bc \neq 0$, we have that the matrix is invertable and
$$\begin{bmatrix}
x_n\\
y_n
\end{bmatrix} = \begin{bmatrix}
a&b\\
c&d
\end{bmatrix}^{-1} \begin{bmatrix}
u_n\\
v_n
\end{bmatrix}$$
So $\begin{bmatrix}
u_n\\
v_n
\end{bmatrix}$ and $\begin{bmatrix}
x_n\\
y_n
\end{bmatrix}$ are equiconvergent.
Note: This way it's easier to see why do we need the $ad-bc \neq 0$ condition.
A: If$$\lim_{n\to\infty}ax_n+by_n=L\text{ and }\lim_{n\to\infty}cx_n+dy_n=M,$$then$$\lim_{n\to\infty}d(ax_n+by_n)-b(cx_n+dy_n)=aL-bM.$$But, since $ad-bc=1$ and since $d(ax_n+by_n)-b(cx_n+dy_n)=(ad-bc)x_n$, this proves that$$\lim_{n\to\infty}x_n=dL-bM.$$A similar argument shows that$$\lim_{n\to\infty}y_n=-cL+aM.$$
