To solve the system of equations:
- \( 6x - 5y = 27 \) (Equation 1)
- \( 3x + 5y = 36 \) (Equation 2)
We can use the method of substitution or elimination. Here, we'll use the elimination method.
First, let's add the two equations to eliminate \( y \):
-
From Equation 1, we have:
\( 6x - 5y = 27 \)
-
From Equation 2, we have:
\( 3x + 5y = 36 \)
Next, we can add these two equations together:
\[ (6x - 5y) + (3x + 5y) = 27 + 36 \]
This simplifies to:
\[ 6x + 3x = 63 \]
\[ 9x = 63 \]
Now, divide both sides by 9:
\[ x = 7 \]
Next, we can substitute \( x = 7 \) back into one of the original equations to find \( y \). We'll use Equation 2:
\[ 3(7) + 5y = 36 \]
This simplifies to:
\[ 21 + 5y = 36 \]
Now, subtract 21 from both sides:
\[ 5y = 15 \]
Dividing both sides by 5 gives:
\[ y = 3 \]
So the solution to the system of equations is:
\[ (x, y) = (7, 3) \]
Thus, the solution is:
\[ \boxed{(7, 3)} \]