Let's solve the system of equations using the addition method. The two equations are:

1. $x - 2y = -3$
2. $-2x + 4y = 6$

Step 1: Add the equations directly.

Notice that if we add the two equations as they are, we can eliminate $x$ because the coefficients of $x$ in the two equations are opposites:

$(x - 2y) + (-2x + 4y) = -3 + 6$

Simplifying:

$x - 2x - 2y + 4y = 3$

This reduces to:

$- x + 2y = 3$

Step 2: Solve for $x$.

The simplified equation is:

$- x + 2y = 3$

We can solve for $x$ by adding $2y$ to both sides:

$x = 2y - 3$

Step 3: Substitute $x = 2y - 3$ into one of the original equations.

Now substitute $x = 2y - 3$ into the first equation, $x - 2y = -3$:

$(2y - 3) - 2y = -3$

Simplifying:

$2y - 3 - 2y = -3$

This simplifies to:

$-3 = -3$

Step 4: Interpret the result.

Since this is a true statement, it means that the system has infinitely many solutions. The equations are dependent (they represent the same line), and every point on that line is a solution.

Final Answer: Infinitely many solutions.

The solution is a parametric solution where $x = 2y - 3$, with $y$ as a free parameter.

---

Would you like more details or have any questions?

Here are some related problems for further practice:

1. Solve the system using substitution: $x + y = 4$ and $2x - y = 1$.
2. Solve the system using elimination: $3x + 2y = 6$ and $2x - 2y = 4$.
3. Check if the system $x + y = 5$ and $2x + 2y = 10$ has infinitely many solutions.
4. Solve $x - y = 7$ and $x + 2y = 3$ using the addition method.
5. Find the intersection point of $x + 3y = 9$ and $2x - y = 4$.

Tip: If you obtain a true statement like $0 = 0$ or $-3 = -3$ after eliminating variables, it typically means the system has infinitely many solutions.