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.

Question

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.

Answer 1

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.