To solve the system of equations:
- \( x + 3y = 15 \) \quad (Equation 1)
- \( 3x + 2y = 10 \) \quad (Equation 2)
We can use the substitution or elimination method. Here, I will use substitution.
From Equation 1, isolate \( x \):
\[ x = 15 - 3y \]
Now, substitute this expression for \( x \) into Equation 2:
\[ 3(15 - 3y) + 2y = 10 \]
Distribute the 3:
\[ 45 - 9y + 2y = 10 \]
Combine the terms with \( y \):
\[ 45 - 7y = 10 \]
Now, isolate \( y \):
\[ -7y = 10 - 45 \] \[ -7y = -35 \] \[ y = \frac{-35}{-7} \] \[ y = 5 \]
Thus, the value of \( y \) for the solution to the system is:
\[ y = 5 \]