To solve the system of equations given:
- \( x + 3y = 15 \) (Equation 1)
- \( 4x + 2y = 30 \) (Equation 2)
Let's start with Equation 1 and express \( x \) in terms of \( y \):
\[ x = 15 - 3y \]
Next, we will substitute this expression for \( x \) into Equation 2.
Substituting \( x \):
\[ 4(15 - 3y) + 2y = 30 \]
Now distribute \( 4 \):
\[ 60 - 12y + 2y = 30 \]
Combine like terms:
\[ 60 - 10y = 30 \]
Subtract \( 60 \) from both sides:
\[ -10y = 30 - 60 \] \[ -10y = -30 \]
Divide by \( -10 \):
\[ y = 3 \]
Now that we have \( y \), substitute it back into the expression for \( x \):
\[ x = 15 - 3(3) \] \[ x = 15 - 9 \] \[ x = 6 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (6, 3) \]
The correct answer is (6, 3).