` 3 x + 5y = 12, 5x + 3y = 4`.

Answer

To solve the system of equations given by:

1. \( 3x + 5y = 12 \) (Equation 1)
2. \( 5x + 3y = 4 \) (Equation 2)

we can use the method of elimination. Here’s a step-by-step solution:

### Step 1: Multiply the equations to align coefficients
To eliminate one of the variables, we can multiply the equations by suitable numbers. We will multiply Equation 1 by 5 and Equation 2 by 3 so that the coefficients of \(x\) will be the same.

- Multiply Equation 1 by 5:
\[ 5(3x + 5y) = 5(12) \implies 15x + 25y = 60 \quad \text{(Equation 3)} \]

- Multiply Equation 2 by 3:
\[ 3(5x + 3y) = 3(4) \implies 15x + 9y = 12 \quad \text{(Equation 4)} \]

### Step 2: Subtract the two new equations
Now, we will subtract Equation 4 from Equation 3 to eliminate \(x\):

\[ (15x + 25y) - (15x + 9y) = 60 - 12 \]

This simplifies to:
\[ 25y - 9y = 48 \]
\[ 16y = 48 \]

### Step 3: Solve for \(y\)
Now, we can solve for \(y\):
\[ y = \frac{48}{16} = 3 \]

### Step 4: Substitute \(y\) back into one of the original equations
Now that we have \(y\), we can substitute \(y = 3\) back into either of the original equations. We will use Equation 1:

\[ 3x + 5(3) = 12 \]
\[ 3x + 15 = 12 \]

### Step 5: Solve for \(x\)
Now, we will isolate \(x\):
\[ 3x = 12 - 15 \]
\[ 3x = -3 \]
\[ x = \frac{-3}{3} = -1 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = -1, \quad y = 3 \]