$$  $$\begin{aligned}

& 13x-6y = 22

\\\\

& x=y+6

\end{aligned}

We have the system of equations:

\[
\begin{aligned}
& 13x - 6y = 22 \quad \text{(Equation 1)} \\
& x = y + 6 \quad \text{(Equation 2)}
\end{aligned}
\]

### Step 1: Substitute $ x $ from Equation 2 into Equation 1

Since $ x = y + 6 $, we replace $ x $ in Equation 1:

\[
13(y + 6) - 6y = 22
\]

### Step 2: Expand and Solve for $ y $

\[
13y + 78 - 6y = 22
\]

\[
7y + 78 = 22
\]

\[
7y = 22 - 78
\]

\[
7y = -56
\]

\[
y = -8
\]

### Step 3: Solve for $ x $

Substituting $ y = -8 $ into Equation 2:

\[
x = -8 + 6
\]

\[
x = -2
\]

### Final Answer:
\[
\boxed{(-2, -8)}
\]

Would you like a detailed explanation of any step? 😊

---

### Related Questions:
1. How do you solve a system of equations using the elimination method?
2. Can this system be solved graphically? What would the intersection point be?
3. How can substitution be used to solve nonlinear systems?
4. What are real-world applications of solving systems of equations?
5. How do you determine if a system of equations has no solution or infinite solutions?

### **Tip:**  
When substituting, always double-check your replacement step to avoid calculation mistakes. ✅

Solve the system of equations: 13x - 6y = 22 and x = y + 6.

Learn how to solve the system of equations 13x - 6y = 22 and x = y + 6 using the substitution method. This step-by-step solution explains key algebraic concepts and demonstrates the correct approach to finding the intersection point (-2, -8). Suitable for students in Grades 8-10.

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