First of all, to solve this with the simplex method (tableau method) the inequalities of the contraints should be equalities. So you have to use slack variables.
$$4x_1+3x_2 \leq 600$$
$$x_1+x_2 \leq 160$$
$$3x_1+7x_2 \leq 840$$
$$x_1,x_2 \geq 0$$
$$\Downarrow$$
$$4x_1+3x_2 +x_3 = 600$$
$$x_1+x_2 +x_4 =160$$
$$3x_1+7x_2 +x_5= 840$$
$$x_1,x_2,x_3, x_4, x_5 \geq 0$$
Then can you continue by creating the tableau?
As regards the Algebraic method:
Draw the $x_1$-axis and the $x_2$-axis. Then from the first contraint you get:
$4x_1+3x_2 \leq 600 : \text{ when } x_1=0 \Rightarrow x_2=200 \text{ and when } x_2=0 \Rightarrow x_1=150$. So you get the line as below, and since it is $\leq 600$ the feasible region is below this line.
Can you do this for the other contraints and find the solution? You should also keep in mind that $x_1, x_2 \geq 0$
Now, I see how to establish an initial tableau for LP problems with >= or = constraints. The following 2 videos are really helpful!
How to Solve a Linear Programming Problem Using the Big M Method
How to Solve a Linear Programming Problem Using the Two Phase Method
Take the LP problem in my question and Big M method for example. We need first convert it to standard form, which is given as follow:
\begin{align}
&\max & v = x_3 - Ma \\
&s.t. & -5x_1 - 3x_2 + x_3 + s_1 = 0\\
& & -2x_1 - 6x_2 + x_3 + s_2 = 0\\
& & -4x_1 - 5x_2 + x_3 + s_3 = 0\\
& & x_1 + x_2 + a = 1 \\
& & x_1, x_2, s_1, s_2, s_3, a \geq 0
\end{align}
where $s_1,s_2,s_3$ are slack variables and $a$ is the artificial variable.
Rewriting $v=x_3-Ma$ as $v-x_3+Ma=0$, we can establish the following initial tableau:
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
Basis & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & a & RHS \\ \hline
s_1 & -5 & -3 & 1 & 1 & 0 & 0 & -1 & 0 \\ \hline
s_2 & -2 & -6 & 1 & 0 & 1 & 0 & 0 & 0 \\ \hline
s_3 & -4 & -5 & 1 & 0 & 0 & 1 & 0 & 0 \\ \hline
a & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 \\ \hline
\tilde{v}& 0 & 0 & -1 & 0 & 0 & 0 & M & 0 \\ \hline
v & -M & -M & -1 & 0 & 0 & 0 & 0 & -M \\ \hline
\end{array}
Note that, the $\tilde{v}$ row is temporary and not a part as the initial tableau, as it needs to be pivoted (the coefficient of $a$ is not $0$).
Then, we can see from the above tableau, the entering variable can be $x_1$, and the departing variable can be $a$. After pivoting, we get the following tableau,
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
Basis & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & a & RHS \\ \hline
s_1 & 0 & 2 & 1 & 1 & 0 & 0 & 5 & 5 \\ \hline
s_2 & 0 & -4 & 1 & 0 & 1 & 0 & 2 & 2 \\ \hline
s_3 & 0 & -1 & 1 & 0 & 0 & 1 & 0 & 4 \\ \hline
x_1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 \\ \hline
v & 0 & 0 & -1 & 0 & 0 & 0 & M & 0 \\ \hline
\end{array}
The next entering variable is $x_3$ and the departing variable is $s_2$. After another pivoting, we get the following tableau,
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
Basis & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & a & RHS \\ \hline
s_1 & 0 & 6 & 0 & 1 & 0 & 0 & 3 & 3 \\ \hline
x_3 & 0 & -4 & 1 & 0 & 1 & 0 & 2 & 2 \\ \hline
s_3 & 0 & 3 & 0 & 0 & -1 & 1 & -2 & 2 \\ \hline
x_1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 \\ \hline
v & 0 & -4 & 0 & 0 & 1 & 0 & M+2& 2 \\ \hline
\end{array}
Now, the entering variable is $x_2$ and the deparing varible is $s_1$. Thus, we can get the next-step tableau,
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
Basis &x_1&x_2&x_3& s_1 & s_2 & s_3 & a & RHS \\ \hline
x_2 & 0 &1 &0 &\frac{1}{6}& 0 & 0 &\frac{1}{2}&\frac{1}{2}\\ \hline
x_3 & 0 & 0 & 1 &\frac{2}{3}& 1 & 0 & 4 & 4 \\ \hline
s_3 & 0 & 0 & 0 &-\frac{1}{2}& -1 & 1 &-\frac{5}{2}&\frac{1}{2}\\ \hline
x_1 & 1 & 0 & 0 &-\frac{1}{6}& 0 & 0 &\frac{1}{2}&\frac{1}{2}\\ \hline
v & 0 & 0 & 0 &\frac{2}{3}& 1 & 0 & M+2 & 4 \\ \hline
\end{array}
As all the coefficients in $v$ row are positive now and $a$ is not in the basis, we get the optimal solution for the original LP problem. That is,
$$x_1=\frac{1}{2}, x_2=\frac{1}{2}, v=4.$$
Best Answer
Find the optimal solution by the dual simplex algorithm
Find the optimal solution by GNU Octave
I post the Octave code for you to verify the optimal solution shown above.