To solve this system of equations:

1. First equation: $y = 2x + 6$
2. Second equation: $3x + 2y = -9$

We will solve using substitution since $y$ is already expressed in terms of $x$ in the first equation.

Step 1: Substitute $y = 2x + 6$ into the second equation.

The second equation is: $3x + 2y = -9$ Replace $y$ with $2x + 6$: $3x + 2(2x + 6) = -9$

Step 2: Simplify the equation.

Distribute $2$ into $(2x + 6)$: $3x + 4x + 12 = -9$

Combine like terms: $7x + 12 = -9$

Step 3: Solve for $x$.

Subtract 12 from both sides: $7x = -21$

Divide by 7: $x = -3$

Step 4: Solve for $y$.

Now that we know $x = -3$, substitute it into the first equation $y = 2x + 6$: $y = 2(-3) + 6$

Simplify: $y = -6 + 6 = 0$

Final Solution:

The solution to the system of equations is: $x = -3, \quad y = 0$

Verification:

Substitute $x = -3$ and $y = 0$ into the second equation $3x + 2y = -9$: $3(-3) + 2(0) = -9$ Simplifies to: $-9 = -9$ This verifies the solution.

Do you want further clarification on any step? Let me know!

Here are 5 related questions:

1. What is the graphical interpretation of this solution for two lines?
2. How would you solve the system using elimination instead of substitution?
3. What happens if two equations in a system have no solution?
4. Can you solve a system of 3 equations with 3 unknowns using substitution?
5. How do you identify dependent systems of equations?

Tip: In a substitution method, always isolate one variable first to simplify the process.