The system of equations given in the image is:

1. $y = 2x - 5$
2. $y = -x + 1$

You are asked to solve this system graphically. To do this, let's find the points where the two lines intersect.

Step 1: Graph the first equation $y = 2x - 5$

This is a straight line with:

• Slope $m = 2$ (meaning it rises 2 units for every 1 unit increase in $x$).
• Y-intercept at $(0, -5)$.

You can plot this line by starting at $(0, -5)$ and using the slope to move up 2 units and 1 unit to the right to plot additional points.

Step 2: Graph the second equation $y = -x + 1$

This is also a straight line with:

• Slope $m = -1$ (meaning it falls 1 unit for every 1 unit increase in $x$).
• Y-intercept at $(0, 1)$.

Plot this line starting at $(0, 1)$ and moving down 1 unit and 1 unit to the right to mark more points.

Step 3: Find the Intersection Point

The intersection of these two lines gives the solution to the system of equations. Solving algebraically:

Set $2x - 5 = -x + 1$.

$2x + x = 1 + 5$ $3x = 6 \quad \Rightarrow \quad x = 2$

Substitute $x = 2$ into one of the original equations, say $y = 2x - 5$:

$y = 2(2) - 5 = 4 - 5 = -1$

Thus, the solution is $x = 2$, $y = -1$. The lines intersect at the point $(2, -1)$.

Do you want more details or have any further questions? Here are some related questions to deepen your understanding:

1. How can you verify if the solution $(2, -1)$ satisfies both equations?
2. What happens if two lines are parallel in a system of equations? How is the solution affected?
3. Can you solve this system using substitution or elimination, and how does the solution compare?
4. What does the slope of a line tell you about how it behaves on a graph?
5. How would the graph look if the lines were perpendicular?

Tip: When graphing lines, always pay attention to the slope and y-intercept, as they determine the position and steepness of the line.