To solve the system of equations:
- \( 6x + 3y = 33 \) (Equation 1)
- \( 5x - 8y = -25 \) (Equation 2)
We can use the substitution or elimination method. Here, we will use the elimination method.
First, let's simplify Equation 1 by dividing everything by 3:
\[ 2x + y = 11 \quad (Equation ; 1') \]
Now, we can express \( y \) in terms of \( x \):
\[ y = 11 - 2x \quad (Equation ; 3) \]
Now, we can substitute this expression for \( y \) into Equation 2:
\[ 5x - 8(11 - 2x) = -25 \] \[ 5x - 88 + 16x = -25 \] \[ 21x - 88 = -25 \] \[ 21x = -25 + 88 \] \[ 21x = 63 \] \[ x = 3 \]
Now that we have \( x \), we can substitute it back into Equation 3 to find \( y \):
\[ y = 11 - 2(3) \] \[ y = 11 - 6 \] \[ y = 5 \]
The solution to the system of equations is:
\[ (x, y) = (3, 5) \]
To verify, we can plug \( x = 3 \) and \( y = 5 \) back into the original equations:
-
For Equation 1: \[ 6(3) + 3(5) = 18 + 15 = 33 \quad \text{(True)} \]
-
For Equation 2: \[ 5(3) - 8(5) = 15 - 40 = -25 \quad \text{(True)} \]
Both equations are satisfied, confirming our solution. Thus, the final answer is:
\[ \boxed{(3, 5)} \]