To solve the system of equations using substitution, we'll follow these steps systematically. The given equations are:
- \( 2x + 2y = 38 \) (Equation 1)
- \( y = x + 3 \) (Equation 2)
Step 1: Substitute Equation 2 into Equation 1
Since Equation 2 expresses \( y \) in terms of \( x \), we can substitute \( y \) in Equation 1:
\[ 2x + 2(x + 3) = 38 \]
Step 2: Simplify the equation
Now, distribute the \( 2 \) on the left side:
\[ 2x + 2x + 6 = 38 \]
Combine like terms:
\[ 4x + 6 = 38 \]
Step 3: Solve for \( x \)
Next, subtract \( 6 \) from both sides:
\[ 4x = 38 - 6 \]
\[ 4x = 32 \]
Now, divide by \( 4 \):
\[ x = \frac{32}{4} = 8 \]
Step 4: Substitute back to find \( y \)
Now that we have \( x \), we can use Equation 2 to find \( y \):
\[ y = x + 3 = 8 + 3 = 11 \]
Step 5: State the solution
The solution to the system of equations is:
\[ x = 8, \quad y = 11 \]
Step 6: Write the final answer
Thus, the solution in coordinate form is:
\[ (8, 11) \]
Verification
To verify, we can insert \( x = 8 \) and \( y = 11 \) back into the original equations:
- For \( 2x + 2y = 38 \):
\[ 2(8) + 2(11) = 16 + 22 = 38 \quad \text{(True)} \]
- For \( y = x + 3 \):
\[ 11 = 8 + 3 \quad \text{(True)} \]
Both equations check out. Therefore, our solution \( (8, 11) \) is correct.