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.

Step-by-step solution to the system of linear equations x - 2y + 4 = 0 and 2x + y + 3 = 0 using the substitution method. Understand how to solve for one variable in terms of another and find values for x and y. Ideal for students in Grades 8-10.

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