I need to find the solutions of the simultaneous congruences $$ \begin{aligned} 3 x+3 z & \equiv 1\pmod 5, \\ 4 x-y+z & \equiv 3\pmod 5 . \end{aligned} $$
How to reduce this into the following ?
\begin{aligned} &x \equiv u+2 v \equiv 2 v+2&\pmod 5, \\ &y \equiv v=v &\pmod 5, \\ &z \equiv t+u+3 v \equiv 3 v &\pmod 5 \\ & \end{aligned}
We are given :
$$\begin{aligned} (1) \ \ \ \ \ & 3 x + 0 y + 3 z & \equiv 1\pmod 5 , \\ (2) \ \ \ \ \ & 4 x - 1 y + 1 z & \equiv 3\pmod 5 . \end{aligned}$$
$$\begin{aligned} 4 \times (1) \ \ \ \ \ & 12 x + 0 y + 12 z & \equiv 4\pmod 5 , \\ 3 \times (2) \ \ \ \ \ & 12 x - 3 y + \ 3 z & \equiv 9\pmod 5 . \end{aligned}$$
$$\begin{aligned} (3) \ \ \ \ \ & 0 x + 3 y + 9 z & \equiv -5\pmod 5 , \\ (3) \ \ \ \ \ & 0 x + 6 y + 8 z & \equiv 0\pmod 5 , \\ (3) \ \ \ \ \ & 0 x + 1 y + 3 z & \equiv 0\pmod 5 , \\ \end{aligned}$$
$$\begin{aligned} 1 \times (1) \ \ \ \ \ & 3 x + 0 y + 3 z & \equiv 1\pmod 5 , \\ 3 \times (2) \ \ \ \ \ & 12 x - 3 y + 3 z & \equiv 9\pmod 5 . \end{aligned}$$
$$\begin{aligned} (4) \ \ \ \ \ & -9 x + 3 y + 0 z & \equiv -8\pmod 5 , \\ (4) \ \ \ \ \ & 1 x + 3 y + 0 z & \equiv 2\pmod 5 , \\ \end{aligned}$$
$$\begin{aligned} (3) \ \ \ \ \ & y + 3 z & \equiv 0\pmod 5 , \\ (3) \ \ \ \ \ & 2 y + 6 z & \equiv 0\pmod 5 , \\ (3) \ \ \ \ \ & 2 y + z & \equiv 0\pmod 5 , \\ (5) \ \ \ \ \ & z & \equiv -2y\pmod 5 , \\ ( \star 5 \star ) \ \ \ \ \ & z & \equiv 3y\pmod 5 , \\ \end{aligned}$$
$$\begin{aligned} (4) \ \ \ \ \ & 1 x + 3 y & \equiv 2\pmod 5 , \\ (6) \ \ \ \ \ & x & \equiv 2-3y\pmod 5 , \\ ( \star 6 \star ) \ \ \ \ \ & x & \equiv 2+2y\pmod 5 , \\ \end{aligned}$$
Hence , the Solution can be $(2y+2,y,3y)$ which is what OP wants to Derive.
We can verify that by using it in the given Equations.