To solve the system of equations using substitution, we start with the following equations:
- \( 6c - 9d = 111 \) (Equation 1)
- \( 5c - 9d = 103 \) (Equation 2)
First, we'll isolate one of the variables in one of the equations. Let's isolate \(d\) in terms of \(c\) using Equation 2.
From Equation 2:
\[ 5c - 9d = 103 \]
Rearranging gives:
\[ 9d = 5c - 103 \] \[ d = \frac{5c - 103}{9} \quad (Equation 3) \]
Now that we have \(d\) in terms of \(c\), we can substitute this expression for \(d\) into Equation 1:
From Equation 1:
\[ 6c - 9d = 111 \]
Substituting for \(d\) from Equation 3:
\[ 6c - 9\left(\frac{5c - 103}{9}\right) = 111 \]
The \(9\) in the denominator and the \(9\) in front cancel out:
\[ 6c - (5c - 103) = 111 \]
This simplifies to:
\[ 6c - 5c + 103 = 111 \] \[ c + 103 = 111 \]
Next, we isolate \(c\):
\[ c = 111 - 103 \] \[ c = 8 \]
Now that we have the value of \(c\), we can substitute it back into Equation 3 to find \(d\):
\[ d = \frac{5(8) - 103}{9} \] \[ d = \frac{40 - 103}{9} \] \[ d = \frac{-63}{9} \] \[ d = -7 \]
Thus, we have found \(c\) and \(d\):
\[ c = 8 \quad \text{and} \quad d = -7 \]
Finally, we present the solution in ordered pair form:
\[ (c, d) = (8, -7) \]
Therefore, the solution to the system of equations is:
\[ \boxed{(8, -7)} \]