Let's solve the given system of equations:

1. $x - 2y + 4 = 0$
2. $2x + y + 3 = 0$

Step 1: Solve for one variable

Let's solve the first equation for $x$ in terms of $y$:

$x = 2y - 4$

Step 2: Substitute $x$ into the second equation

Now, substitute $x = 2y - 4$ into the second equation:

$2(2y - 4) + y + 3 = 0$

Expanding and simplifying:

$4y - 8 + y + 3 = 0$ $5y - 5 = 0$ $y = 1$

Step 3: Substitute $y = 1$ back to find $x$

Now that we know $y = 1$, substitute this value into $x = 2y - 4$:

$x = 2(1) - 4 = 2 - 4 = -2$

Solution

The solution to the system is: $x = -2, \quad y = 1$

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. How can you verify that the solution is correct for both equations?
2. What happens if you try to solve for $y$ first instead of $x$?
3. How do you approach solving a system of linear equations with three variables?
4. Can systems of equations have no solutions? If so, why?
5. What is the graphical interpretation of the solution to this system?

Tip: When solving systems of equations, substitution is often easiest when one equation can be rearranged to isolate a variable with a coefficient of 1 or -1.