Consider the following linear problem
$$\max tx_1+x_2\\ s.t. 4x_1+3x_2\le12 \\ 3x_1+4x_2\le12\\ x_1,x_2\ge0$$
where the parameter $t$ grows exponentially $t\in[1,\infty).$
Find the sequence of basic optimal solutions.
Attempt:
First we should convert to standar form
$$\max tx_1+x_2\\ s.t.\ 4x_1+3x_2+x_3=12 \\ 3x_1+4x_2+x_4=12\\ x_1,x_2\ge0$$
Then, I am not sure how should I proceed.
I've solved exercise where you have say $3$ instead of $t$ and then you change $3$ by $4$ and then analyze how the final simplex tableau changes, as it is a coefficient of a basic variable the optimality and factibility might both be affected.
But this could be fixed by using again simplex method or dual-simplex in the final tableau.
My question is how should I proceed in this case where there is a $t$ instead of a number?
Could someone help me please?
I would really appreciate any help you're willing to provide. Thank you.

You can work geometrically. The domain is given in the picture below :
Now, the gradient of $f(x_1,x_2) = tx_1+x_2$ equals $\pmatrix{t\\1}$.
If $t=1$, the optimum lies at the intersection of $3x_1+4x_2=12$ and $4x_1+3x_2=12$, i.e. $(\frac{12}{7},\frac{12}{7})$.
If $t=\frac{4}{3}$, the gradient is orthogonal to $4x_1+3x_2=12$, which means that all points on this line are optimal, in particular $(\frac{12}{7},\frac{12}{7})$ and $(3,0)$.
So for any $t\in [1,\frac{4}{3}[$, $(\frac{12}{7},\frac{12}{7})$ is the optimal point.
Then, as $t$ increases, the gradient points towards $(3,0)$.
Note that this matches gt6989b's answer : solving $3t = \frac{12(t+1)}{7}$ for $t$ yields $t=\frac{4}{3}$.