Question

y = -\frac{1}{2}x + 3
click to select points on the graph.
x + 2y = 6
y = -\frac{1}{2}x + 3

Explanation:

Step1: Analyze the equations

We have two equations: \( x + 2y = 6 \) and \( y = -\frac{1}{2}x + 3 \). Let's check if they are equivalent.
From \( y = -\frac{1}{2}x + 3 \), multiply both sides by 2: \( 2y = -x + 6 \), then add \( x \) to both sides: \( x + 2y = 6 \). So the two equations represent the same line.

Step2: Identify points on the line

Looking at the graph, we can see the y - intercept (when \( x = 0 \)): substitute \( x = 0 \) into \( y = -\frac{1}{2}x + 3 \), we get \( y = 3 \), so the point is \( (0, 3) \).
The x - intercept (when \( y = 0 \)): substitute \( y = 0 \) into \( y = -\frac{1}{2}x + 3 \), we have \( 0 = -\frac{1}{2}x + 3 \), then \( \frac{1}{2}x = 3 \), so \( x = 6 \), and the point is \( (6, 0) \). Also, we can check other points. For example, if we take \( x=-10 \), then \( y = -\frac{1}{2}(-10)+3=5 + 3=8 \), so the point \( (-10, 8) \) is on the line, and when \( x = 10 \), \( y=-\frac{1}{2}(10)+3=-5 + 3=-2 \), so the point \( (10, -2) \) is on the line.

Answer:

The two equations \( x + 2y = 6 \) and \( y = -\frac{1}{2}x + 3 \) are equivalent (represent the same line). Points on the line include \( (0, 3) \), \( (6, 0) \), \( (-10, 8) \), \( (10, -2) \) etc. (If the question was to show the relationship between the equations or identify points, this is the relevant information. Since the original question was not fully specified, but based on the graph and equations, we can conclude the equations are the same and identify key points on the line.)