To solve the system of equations from the image:

1. **Equations:**
\[
12x + 5y = 12 \quad (1)
\]
\[
2x + 3y = 4 \quad (2)
\]

2. **Simplify equations**:
Let's first reduce equation (1) by dividing everything by 2:
\[
6x + \frac{5}{2}y = 6 \quad (3)
\]

3. **Now, we'll use equation (2) (as is) and equation (3)** to eliminate \(y\).

4. **Solve equation (2) for \(y\)**:
\[
3y = 4 - 2x \implies y = \frac{4 - 2x}{3} \quad (4)
\]

5. **Substitute equation (4) into equation (3)**:
\[
6x + \frac{5}{2}\left(\frac{4 - 2x}{3}\right) = 6
\]

6. **Multiply through by 6 to eliminate the fraction**:
\[
36x + 5(4 - 2x) = 36
\]
\[
36x + 20 - 10x = 36
\]
\[
26x + 20 = 36
\]
\[
26x = 16 \implies x = \frac{16}{26} = \frac{8}{13}
\]

7. **Substitute \(x\) back into equation (4)** for \(y\):
\[
y = \frac{4 - 2\left(\frac{8}{13}\right)}{3} = \frac{4 - \frac{16}{13}}{3}
\]
\[
= \frac{\frac{52 - 16}{13}}{3} = \frac{\frac{36}{13}}{3} = \frac{36}{39} = \frac{12}{13}
\]

8. **Final Solutions**:
\[
x = \frac{8}{13}, \quad y = \frac{12}{13}
\]

These are the values of \(x\) and \(y\) that satisfy the original equations.