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We are given the following system of equations:

\[
\begin{aligned}
y &= \frac{1}{2}x + 1 \\
x - 2y &= -2
\end{aligned}
\]

### Step 1: Solve the system algebraically

To solve the system, we can use substitution or elimination. I'll start by using **substitution** since the first equation is already solved for \( y \).

1. From the first equation: 
   \[
   y = \frac{1}{2}x + 1
   \]
   
2. Substitute this expression for \( y \) into the second equation:
   \[
   x - 2\left(\frac{1}{2}x + 1\right) = -2
   \]

### Step 2: Simplify and solve for \( x \)

Expand and simplify the equation:
\[
x - 2\left(\frac{1}{2}x + 1\right) = -2
\]
\[
x - x - 2 = -2
\]
\[
-2 = -2
\]

This simplifies to a **true statement** \( -2 = -2 \), meaning that the two lines are actually the same line. Therefore, this system has **infinitely many solutions**.

### Step 3: Graph the system

Since both equations represent the same line, graphing them would show a single line, and every point on this line is a solution to the system.

### Solution

The system has **infinitely many solutions** because the two equations describe the same line.

---

Would you like more details or have any questions?

Here are five related questions:
1. How do you graph linear equations on a coordinate plane?
2. What does it mean when two lines are the same on a graph?
3. How do you check for parallel or coincident lines in a system of equations?
4. What is the difference between substitution and elimination methods?
5. How can you determine if a system of equations has no solution?

**Tip**: Whenever you get a true statement like \( 0 = 0 \) or \( -2 = -2 \), it indicates that the system has infinitely many solutions!

Graph the system of equations: y = (1/2)x + 1 and x - 2y = -2, and write its solution.

Learn how to graph and solve a system of linear equations involving y = (1/2)x + 1 and x - 2y = -2. This solution shows that the system has infinitely many solutions because the two equations represent the same line. Explore key algebraic concepts such as the substitution method and coincident lines.

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