Question

Use the image to answer the question.

A coordinate graph goes from negative 10 to 10 in 1-unit increments on both the x- and y-axes. Two linear equations are graphed, along with two points.

The image shows the graph of the system of linear equations {y=2x+2y=2x−4
. Which of the following points, if any, are solutions to the system of equations?

(1 point)
Responses

(0,2)
open paren 0 comma 2 close paren

Both of these points are solutions.
Both of these points are solutions.

Neither of these points are solutions.
Neither of these points are solutions.

(0,−4)

Answer 1

To determine if the points (0, 2) and (0, -4) are solutions to the system of linear equations \(y = 2x + 2\) and \(y = 2x - 4\), we can substitute the \(x\) values of the points into both equations and see if we get the corresponding \(y\) values.

1. **For the point (0, 2):**
- Substitute \(x = 0\) into the first equation:
\[
y = 2(0) + 2 = 2
\]
This matches the \(y\) value of the point (0, 2).
- Substitute \(x = 0\) into the second equation:
\[
y = 2(0) - 4 = -4
\]
This does not match the \(y\) value of the point (0, 2).

Thus, (0, 2) is a solution to the first equation but not to the second.

2. **For the point (0, -4):**
- Substitute \(x = 0\) into the first equation:
\[
y = 2(0) + 2 = 2
\]
This does not match the \(y\) value of the point (0, -4).
- Substitute \(x = 0\) into the second equation:
\[
y = 2(0) - 4 = -4
\]
This matches the \(y\) value of the point (0, -4).

Thus, (0, -4) is a solution to the second equation but not to the first.

Given this analysis:

- The point (0, 2) is a solution to \(y = 2x + 2\), but not to \(y = 2x - 4\).
- The point (0, -4) is a solution to \(y = 2x - 4\), but not to \(y = 2x + 2\).

**Final Answer: Neither of these points are solutions to the system of equations.**