Certainly! Let’s solve the given system of linear equations by graphing:
[tex]\[
\begin{array}{l}
y = -\frac{1}{2} x + 2 \\
y = \frac{1}{2} x - 1
\end{array}
\][/tex]
### Step-by-Step Solution
Step 1: Graph the first equation [tex]\( y = -\frac{1}{2} x + 2 \)[/tex]
1. Find the y-intercept: This is the point where [tex]\( x = 0 \)[/tex].
[tex]\[
y = -\frac{1}{2} \cdot 0 + 2 \implies y = 2
\][/tex]
So, the y-intercept is [tex]\( (0, 2) \)[/tex].
2. Find another point for the line: Choose [tex]\( x = 2 \)[/tex] for easy calculation.
[tex]\[
y = -\frac{1}{2} \cdot 2 + 2 = -1 + 2 = 1
\][/tex]
So, another point on the line is [tex]\( (2, 1) \)[/tex].
3. Plot these points and draw the line:
- Plot [tex]\( (0, 2) \)[/tex]
- Plot [tex]\( (2, 1) \)[/tex]
- Draw the line through these points.
Step 2: Graph the second equation [tex]\( y = \frac{1}{2} x - 1 \)[/tex]
1. Find the y-intercept: This is the point where [tex]\( x = 0 \)[/tex].
[tex]\[
y = \frac{1}{2} \cdot 0 - 1 \implies y = -1
\][/tex]
So, the y-intercept is [tex]\( (0, -1) \)[/tex].
2. Find another point for the line: Choose [tex]\( x = 2 \)[/tex] for easy calculation.
[tex]\[
y = \frac{1}{2} \cdot 2 - 1 = 1 - 1 = 0
\][/tex]
So, another point on the line is [tex]\( (2, 0) \)[/tex].
3. Plot these points and draw the line:
- Plot [tex]\( (0, -1) \)[/tex]
- Plot [tex]\( (2, 0) \)[/tex]
- Draw the line through these points.
Step 3: Find the point of intersection
Once both lines are graphed, the point where they intersect is the solution to the system of equations.
By closely examining the graphs of these two lines, the point where they intersect is [tex]\( (3, 0.5) \)[/tex].
### Conclusion
The solution to the system of equations is:
[tex]\[
\boxed{(3, 0.5)}
\][/tex]
