Question

pretest: systems of linear equations and inequalities
match each point of intersection with the system of equations whose solution is at that point.
(graph with points w, x, y, z on coordinate plane)

Answer:

To solve this, we first identify the coordinates of each point:

• Point W: From the graph, it's at \( x = -2 \), \( y = 2 \)? Wait, no—wait, looking again: Wait, W is at \( x = -2 \), \( y = 2 \)? Wait, no, let's check the grid. Wait, the x-axis: -2 is the x-coordinate, and y-coordinate? Wait, the red dot W: x=-2, y=2? Wait, no, maybe I misread. Wait, the vertical lines are x, horizontal y. Let's re-express:

• Point X: \( (-1, 3) \) (x=-1, y=3)
• Point W: \( (-2, 2) \)? Wait, no, the blue dot W: x=-2, y=2? Wait, maybe the lines: Let's assume we need to find the system of equations for each point. But since the problem is to match points to systems, we first find the coordinates:

1. Find coordinates of each point:

• \( W \): \( x = -2 \), \( y = 2 \) (wait, no, looking at the graph, W is at x=-2, y=2? Wait, the grid: each square is 1 unit. So:
• \( X \): \( (-1, 3) \)
• \( W \): \( (-2, 2) \)
• \( Y \): \( (1, 1) \)
• \( Z \): \( (0, -2) \) (wait, Z is on the y-axis? No, Z is at x=0? Wait, no, Z is at x=-0.5? No, the graph: Z is at x=0? Wait, the red dot Z is on the y-axis? Wait, no, the vertical line through x=0 (y-axis) and y=-2? Wait, maybe the coordinates are:

Let's list each point:

• \( W \): \( (-2, 2) \)
• \( X \): \( (-1, 3) \)
• \( Y \): \( (1, 1) \)
• \( Z \): \( (0, -2) \) (or maybe \( (0, -2) \)? Wait, the line through Z: maybe two lines intersecting at Z.

1. Determine the system of equations for each point:

For a point \( (a, b) \), the system is two lines that intersect at \( (a, b) \). So we need to find the equations of the lines passing through each point.

For example, take point \( X(-1, 3) \):

• One line might have a slope. Let's assume the pink line (going up) has a slope. Let's calculate slope between two points on the pink line: from \( (-4, 0) \) (x-intercept) to \( (0, 4) \) (y-intercept). Slope \( m = \frac{4 - 0}{0 - (-4)} = 1 \). So equation: \( y = x + 4 \).
• Another line through \( X(-1, 3) \): let's take the blue line. From \( X(-1, 3) \) to \( (0, 2) \) (y-intercept). Slope \( m = \frac{2 - 3}{0 - (-1)} = -1 \). Equation: \( y = -x + 2 \).
• Check if \( (-1, 3) \) satisfies both:
• \( y = x + 4 \): \( 3 = -1 + 4 \) → \( 3 = 3 \) (true).
• \( y = -x + 2 \): \( 3 = -(-1) + 2 \) → \( 3 = 3 \) (true). So the system \(

$$\begin{cases} y = x + 4 \\ y = -x + 2 \end{cases}$$

\) has solution \( (-1, 3) \), which is point \( X \).

Similarly, for point \( W(-2, 2) \):

• One line: the blue line (same as above? No, another line). Wait, the dark blue line: from \( (-2, 2) \) to \( (1, 1) \). Slope \( m = \frac{1 - 2}{1 - (-2)} = -\frac{1}{3} \)? No, maybe better to use intercepts.

But since the problem is to match points to systems, the key is to identify the coordinates of each point and then find the system (two equations) that intersect at that point.

However, since the original problem is a matching task, and without the list of systems, we can only explain the process:

• Step 1: Find the coordinates of each intersection point (e.g., \( W(-2, 2) \), \( X(-1, 3) \), \( Y(1, 1) \), \( Z(0, -2) \)).
• Step 2: For each system of linear equations, solve it (by substitution, elimination, or graphing) to find its solution (x, y).
• Step 3: Match the solution (x, y) to the point with those coordinates.

If we assume the systems are given (even though not listed here), the process is:

1. Solve each system of linear equations to find its solution (x, y).
2. Identify the point with those coordinates.

For example, if a system is \(

$$\begin{cases} y = x + 4 \\ y = -x + 2 \end{cases}$$

\), solving:

• Add the two equations: \( 2y = 6 \) → \( y = 3 \).
• Substitute \( y = 3 \) into \( y = x + 4 \): \( 3 = x + 4 \) → \( x = -1 \).
• Solution: \( (-1, 3) \), which is point \( X \).

So the key is to solve each system, find (x, y), and match to the point with those coordinates.