Let's solve the given system of equations graphically as instructed in the image. The equations are:

1. $x - 2y = -2$
2. $y = -1$

Step 1: Rewrite the first equation in slope-intercept form ($y = mx + b$).

Starting with $x - 2y = -2$, solve for $y$: $x - 2y = -2 \implies -2y = -x - 2 \implies y = \frac{x}{2} + 1$ So the first equation becomes $y = \frac{1}{2}x + 1$.

Step 2: Plot the two equations.

• For $y = -1$: This is a horizontal line where $y$ is always $-1$. It crosses the $y$-axis at $-1$ and is parallel to the $x$-axis.

• For $y = \frac{1}{2}x + 1$: This is a line with slope $\frac{1}{2}$ (rises 1 unit for every 2 units to the right) and $y$-intercept $1$ (crosses the $y$-axis at $y = 1$).

Step 3: Find the point of intersection.

To determine where the lines intersect, substitute $y = -1$ (from the second equation) into the first equation: $-1 = \frac{1}{2}x + 1$ Solve for $x$: $-1 - 1 = \frac{1}{2}x \implies -2 = \frac{1}{2}x \implies x = -4$ Thus, the lines intersect at $(-4, -1)$.

Step 4: Verify graphically.

Plot both lines on the graph:

1. $y = -1$: A horizontal line through $y = -1$.
2. $y = \frac{1}{2}x + 1$: Passes through $(0, 1)$ and $(2, 2)$.

The intersection point is $(-4, -1)$.

Would you like further clarification or a visualization of the graph? Here are 5 related questions to deepen your understanding:

1. How do you graph lines with different slopes and intercepts?
2. Why does the intersection point represent the solution to a system of equations?
3. What is the significance of parallel lines in a system of equations?
4. How do you solve a system of equations algebraically instead of graphically?
5. Can systems of equations have more than one solution?

Tip: Always verify the intersection point by substituting it into both equations!