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The graphical method is a visual approach to solving systems of equations. It involves drawing straight lines based on the equations given and identifying where they intersect.

To apply this method to the system of equations \(3x-2y=6\) and \(2x-3y=-6\), you would first need to rearrange the equations into slope-intercept form, \(y=mx+b\), where \(m\) represents the slope and \(b\) the y-intercept. For the first equation, this gives us \(y=\frac{3x-6}{2}\) and for the second \(y=\frac{2x+6}{3}\).

After plotting both equations on the graph, the point where the two lines intersect represents the solution to the system. If they intersect at a single point, like in our case at (6,6), the system has a unique solution. If they coincide everywhere, the system has infinitely many solutions. If they never meet, the system has no solutions.

The substitution method is another algebraic technique where you solve one of the equations for one variable, then substitute this expression into the other equation. By doing this, you reduce the system to a single equation with one variable, which can be solved.

Using our given equations \(3x-2y=6\) and \(2x-3y=-6\), you might solve the first one for \(x\) to get \(x=\frac{2y+6}{3}\) and then substitute this expression for \(x\) into the second equation. This would give you a new equation in terms of \(y\) alone. From there, you could find the value of \(y\), and then substitute that back into any original equation to solve for \(x\). This method is particularly useful when one of the equations is already solved for one of the variables.

The elimination method involves adding or subtracting the equations to eliminate one of the variables. You usually need to manipulate the equations first, so the same variable in both equations has the same coefficient (but with opposite signs).

In our example, the problem began by making the coefficients of \(y\) the same in both equations. Multiplying the first equation by 3 and the second by 2 did the trick, leading to \(9x - 6y = 18\) and \(4x - 6y = -12\). Afterward, we subtracted one equation from the other, eliminating \(y\) and leaving a simple equation with only \(x\), which we then solved. The value obtained for \(x\) could be substituted into one of the original equations to find \(y\).

Algebraic solutions encapsulate methods like substitution and elimination. They are systematic and follow a series of algebraic manipulations to find solutions for variables.

For the system \(3x-2y=6\) and \(2x-3y=-6\), we used the elimination method to find an algebraic solution, but we could use substitution, solving one equation for one variable and substituting into the other, just as well. Important in algebraic solutions is the check: we substituted \(x=6\) and \(y=6\) back into the original equations to ensure both were satisfied, confirming that (6,6) is indeed the correct solution.