#### Answer

The system with this solution is
\begin{align*}
3x_1-x_2=& 4\\
x_1-\frac{1}{3}x_2 = &\frac{4}{3}.
\end{align*}
For the solution of the second part of the problem see step-by-step below.

#### Work Step by Step

To find the required system we will need to make two equations just in $x_1$ and $x_2$.
Put $x_1$ instead of $t$ in the second given equation:
$$x_2=3x_1-4$$
which can be rewritten as
$$3x_1-x_2=4$$
and there is our first equation.
To make our second equation express $t$ in terms of $x_2$ from the second equation given in the problem:
$$3t=x_2+4\Rightarrow t=\frac{1}{3}x_2+\frac{4}{3}.$$
Now since $x_1 = t$ we have
$$x_1 = \frac{1}{3}x_2+\frac{4}{3}$$
and putting all the variables on the left this becomes
$$x_1-\frac{1}{3}x_2 = \frac{4}{3}.$$
Now our system reads
\begin{align*}
3x_1-x_2=& 4\\
x_1-\frac{1}{3}x_2 =& \frac{4}{3}.
\end{align*}
Now let us go to the second part of the problem. To show that this is possible we will put $t$ instead of $x_2$ (because we want $x_2 = t$) to the second equation in the system:
$$x_1-\frac{1}{3}t = \frac{4}{3}$$ and this gives
$$x_1 = \frac{1}{3}t+\frac{4}{3}$$ and this is what we required.