Any system of equations can be solved for any of those methods. Generally, using a graph and using a table require a lot of work, since it is not always easy to read the coordinates on a graph (specially when dealing with fractionary solutions), and on a table the issue is the same as with the graphs.
Usually, elimination and substitution methods are used depending on the coefficients of the variables on the system. Whenever there is a coefficient of 1 on some variable, the elimination method can be used very easily. Whenever there is already an isolated variable, the substitution method is prefered.
For the system:
[tex]\begin{gathered} 3x+8y=-4 \\ 2x-4y=16 \end{gathered}[/tex]If we multiply the second equation by 2, we get:
[tex]\begin{gathered} 2(2x-4y)=2(16) \\ \Rightarrow4x-8y=32 \end{gathered}[/tex]So the system can be rewritten as:
[tex]\begin{gathered} 3x+8y=-4 \\ 4x-8y=32 \end{gathered}[/tex]Since the coefficient of y is +8 in one equation and -8 in the other, we can use the elimination method by adding both equations side by side.
For the system:
[tex]\begin{gathered} 6x-y=16 \\ x=4y-5 \end{gathered}[/tex]The second equation has already an isolated variable, x. Then, we can use the substitution method by plugging in the expression for x in the first equation.
For the system:
[tex]\begin{gathered} x+y=19 \\ 3x-2y=-3 \end{gathered}[/tex]Since x and y have a coefficient of 1, we can multiply the first equation by 2, so the variable y would get a coefficient of 2, which will help us apply the elimination method, so when adding both equations, +2y and -2y cancell out.