Problem
Given two independent random variables$Z_1 \sim \mathcal{N}(0,1)$ and $Z_1 \sim \mathcal{N}(0,1)$, what are the marginal distributions for $Y_1$ and $Y_2$ in all three cases?
(1) $Y_1 = Z_1 + 2$, $Y_2 = 3Z_1 + 3Z_2$
(2) $Y_1 = Z_1 + 2$, $Y_2 = 3\sqrt{2}Z_2$
(3) $Y_1 = Z_1 + 2$, $Y_2 = 3\sqrt{2}Z_1$
Context and Attempt
This is not a homework problem, just a question asked in a slide deck. Referring to case (1), it seems like these functions can be re-written to $f_{Z_1}(z_1) = z_1 + 2$ and $f_{Z_1, Z_2}(z_1, z_2) = 3 z_1 + 3 z_2$ (per the notation in this question). Since from the marginal distribution wiki, it seems like the joint distribution must be known, and then the marginal distribution can be obtained by integrating the joint probability distribution as shown below:
$$ \begin{aligned} f_{Z_1}(z_1) &= \int_c^d f(z_1, z_2)\ dz_2, \\ f_{Z_2}(z_2) &= \int_a^b f(z_1, z_2)\ dz_1. \end{aligned} $$
So it seems like a possible solution, at least for case (1) is to perform the following integrals
$$ \begin{aligned} f_{Z_1}(z_1) &= \int_0^1 3z_1 + 3z_2\ dz_2 = \left[3 z_1 z_2 + \frac{3 z_2^2}{2} \right]_0^1 = 3 z_1 + \frac{3}{2}, \\ f_{Z_2}(z_2) &= \int_0^1 3z_1 + 3z_2\ dz_1 = \frac{3}{2} + 3 z_2, \end{aligned} $$
however, this doesn't really seem correct in the context of the given problem, even though it generally aligns with the websites I have linked. Moreover, for case (2) and (3) there is no joint probability distribution function provided, so I wouldn't even know how to proceed.