` 2x + 3y = 8 `,
` x - 2y + 3 = 0 `

Answer

To solve the system of equations:

1. **Write down the equations:**
- First equation: \( 2x + 3y = 8 \) (Equation 1)
- Second equation: \( x - 2y + 3 = 0 \) (Equation 2)

2. **Rearrange the second equation:**
- From Equation 2, we can rearrange it to isolate \( x \):
\[ x - 2y = -3 \implies x = 2y - 3 \]
(This is now our modified Equation 2)

3. **Substitute \( x \) in Equation 1:**
- Substitute \( x = 2y - 3 \) into Equation 1:
\[ 2(2y - 3) + 3y = 8 \]
- Simplifying this gives:
\[ 4y - 6 + 3y = 8 \]
\[ 7y - 6 = 8 \]

4. **Solve for \( y \):**
- Add 6 to both sides:
\[ 7y = 14 \]
- Divide by 7:
\[ y = 2 \]

5. **Substitute \( y \) back to find \( x \):**
- Now substitute \( y = 2 \) back into the modified Equation 2:
\[ x = 2(2) - 3 \]
\[ x = 4 - 3 = 1 \]

6. **Final solution:**
- The solution to the system of equations is:
\[ x = 1, \quad y = 2 \]