To solve the system of equations graphically, you need to graph both lines and find their intersection point. Here's how to approach this:

Step 1: Understand the given equations

The system of equations is:

1. $y = \frac{1}{2}x + 4$
2. $y = -x + 7$

Step 2: Graph the first equation $y = \frac{1}{2}x + 4$

This is a linear equation in slope-intercept form, $y = mx + b$, where:

• $m = \frac{1}{2}$ is the slope.
• $b = 4$ is the y-intercept.

1. Start by plotting the y-intercept at (0, 4).
2. Use the slope $\frac{1}{2}$, which means rise over run: go up 1 unit and right 2 units to plot additional points.
3. Draw a straight line through these points.

Step 3: Graph the second equation $y = -x + 7$

This is also a linear equation in slope-intercept form, $y = mx + b$, where:

• $m = -1$ is the slope.
• $b = 7$ is the y-intercept.

1. Start by plotting the y-intercept at (0, 7).
2. Use the slope $-1$, which means go down 1 unit and right 1 unit to plot more points.
3. Draw a straight line through these points.

Step 4: Find the intersection point

The solution to the system of equations is the point where the two lines intersect. You can either read this directly from the graph or solve algebraically.

By graphing, you'll find the lines intersect at the point (6, 7/2). This means: $x = 6, \quad y = 7/2$

Thus, the solution to the system of equations is: $(x, y) = (6, 3.5)$

Would you like to see more details or clarifications? Let me know if you need help with anything!

Here are 5 related questions to explore:

1. How can we verify the solution algebraically without graphing?
2. What is the significance of the slope in a linear equation?
3. How can you determine the number of solutions a system of linear equations has by graphing?
4. What is the general form of a linear equation, and how do you convert it to slope-intercept form?
5. How do parallel lines appear when graphing a system of equations?

Tip: Always double-check the slope and intercept when graphing, as small errors can lead to incorrect intersection points.