Determining time-invariance of a system I have a question about determining time-invariance of a linear system. We are given this system and we need to determine if it is time-invariant or not:

$$y(t)=\int_{-t}^{\infty}x(-3\tau)d\tau$$

Now after doing the shifting $t-t_o$ you get two different outputs. Now that makes sense, but what confuses me is the variable substition that happens in order to check. I'll post a link to the picture on imgur,sorry for this but I don't know how to do it otherwise.

https://imgur.com/a/CBbe4Sp

Basically my question is why is the substittution $-3τ -t0$ and not just $τ -t0$. Also why are we substituing both the Z1 as well as Z2, isn't the goal to make try to make them equal? Like try to modify one to get the same result as the other?.Maybe someone can give me some insight.Sorry for the bad post I need instructions on how to post. Cheers!
You have a system with the following input-output relation:

$$y(t)=\int_{-t}^{\infty}x(-3\tau)d\tau\tag{1}$$

In order to check whether the system is time-invariant or not, we need to compare the shifted output with the output resulting from a shifted input. The shifted output is

$$y(t-T)=\int_{-(t-T)}^{\infty}x(-3\tau)d\tau\tag{2}$$

Shifting the input means applying an input signal $x_T(t)=x(t-T)$. Note that $x_T(-3\tau)=x(-3\tau-T)$, so the response to the shifted input is

$$y_T(t)=\int_{-t}^{\infty}x(-3\tau-T)d\tau\tag{3}$$

The system is time-invariant if $(3)$ equals $(2)$. So we try to make the integrand in $(3)$ look like the one in $(2)$ by substituting $-3\tau-T$ by $-3\zeta$, which gives

$$y_T(t)=\int_{-(t-T/3)}^{\infty}x(-3\zeta)d\zeta\tag{4}$$

Now $(2)$ and $(4)$ have the same integrand, but we see that the lower integration limits are different. Consequently, $y_T(t)\neq y(t-T)$, hence the system is not time-invariant.