https://file.aiquickdraw.com/mathbot/file/17265000652292m8bstsp.jpg

The given system of equations is:

\[
\text{System A:} 
\begin{cases}
2x - y = -6 \\
-2x + y - 6 = 0
\end{cases}
\]

Let's solve this system to determine whether it has no solution, a unique solution, or infinitely many solutions.

### Step 1: Simplify the second equation
The second equation is:
\[
-2x + y - 6 = 0 \implies y = 2x + 6
\]
Now, we can substitute this expression for \( y \) into the first equation.

### Step 2: Substitution
Substitute \( y = 2x + 6 \) into the first equation \( 2x - y = -6 \):
\[
2x - (2x + 6) = -6
\]
Simplifying:
\[
2x - 2x - 6 = -6 \implies -6 = -6
\]

This is a true statement, which means the system is **dependent** and has **infinitely many solutions**.

### Step 3: Conclusion
Since we have a true statement and no contradictions, the system does not have a unique solution or no solution. Instead, the system has infinitely many solutions, and they satisfy the equation \( y = 2x + 6 \).

So, the correct answer is:
- The system has infinitely many solutions.
- The solutions must satisfy the equation: \( y = 2x + 6 \).

Would you like more details or have any other questions?

### 5 Related Questions:
1. How can we tell if a system of equations is dependent?
2. What is the difference between consistent and inconsistent systems?
3. Can we graphically interpret a system with infinitely many solutions?
4. How would you apply substitution in more complex systems?
5. How does elimination compare to substitution for solving systems?

### Tip:
For systems of linear equations, a dependent system (infinitely many solutions) occurs when the equations represent the same line.

Solve the system of equations: 2x - y = -6 and -2x + y - 6 = 0. Determine if the system has no solution, a unique solution, or infinitely many solutions.

Solve the system of equations: 2x - y = -6 and -2x + y - 6 = 0 using the substitution method. Learn how to identify when a system has infinitely many solutions by checking for dependent systems. Suitable for students in Grades 8-10.

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