Is the system represented by the equation $y(t) = x(2t)$ time invariant? I came across this problem in the text book Signals and Systems - Oppenheim (Example-1.16).

To solve this, I followed the following algorithm (described in the book earlier for a separate problem):

$$\begin{align}
y(t) &amp;=x(2t)\\
y_1(t)&amp;=x_1(2t)\end{align}$$

Let $x_2(t)=x_1(t-t_0)$ and
$y_2(t)=x_2(2t)$

$$\begin{align}
\implies y_2(t) &amp;= x_1(2(t - t_0))\\
&amp;= x_1(2t-2t_0)\end{align}$$

Now, $y_1(t-t_0) = x_1(2(t-t_0)) = x_1(2t - 2t_0)$.

Since $y_2(t)$ and $y_1(t-t_0)$ are equivalent, the system should be time-invariant.

However, the book takes a different(graphical) approach, wherein the time-shifted $y(t)$,  $y(t - t _0)$ is $x(2t - t_0)$ and the resulting system is time-varying.

I would like to understand why is there this inconsistency between the mathematical and the graphical approach.
From your solution: 

I followed the following algorithm:

$$ y(t) =x(2t) $$
$$ y_1(t) = x_1(2t)$$

Let  $$x_2(t) = x_1(t-t_0)  ~~~\text{and}~~~ y_2(t) = x_2(2t) $$

On this following step (time sampling of the shifted argument) you make the usual mistake:

$$\implies y_2(t) = x_1(2(t - t_0))$$ which should instead be :
$$\implies y_2(t) = x_2(2t) = x_1(t - t_0)|_{t=2t} = x_1(2t - t_0)$$

The remaining parts follow as usual to show that the time scaler is a time-varying system...