Question

Complete the statement with an equation that makes the statement true.(1 point)
The system of equations 3x+4y=1
and y= ( )
has an infinite number of solutions.

Answer 1

Cruz wrote a system of equations with an infinite number of solutions. One of his equations is 6x+3y=9
. What could be his second equation? (Write the equation in slope-intercept form.)(1 point)
The second equation could be y= ( )

Answer 2

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,−4)
open paren 0 comma negative 4 close paren

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

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

(0,2)

Answer 3

To create an equation for \( y \) that makes the system of equations have an infinite number of solutions with the equation \( 3x + 4y = 1 \), we need the second equation to represent the same line as the first.

To do this, we can rearrange the first equation to solve for \( y \):

1. Start with the equation:
\[
3x + 4y = 1
\]

2. Isolate \( 4y \):
\[
4y = 1 - 3x
\]

3. Divide by 4:
\[
y = \frac{1}{4} - \frac{3}{4}x
\]

Thus, we can complete the statement with:

\[
y = \frac{1}{4} - \frac{3}{4}x
\]

This equation describes the same line as \( 3x + 4y = 1 \) and therefore ensures that the system has an infinite number of solutions.

Answer 4

To ensure that the second equation has an infinite number of solutions with the equation \( 6x + 3y = 9 \), it must represent the same line.

First, let's rewrite the given equation in slope-intercept form (i.e., \( y = mx + b \)).

1. Start with the equation:
\[
6x + 3y = 9
\]

2. Isolate \( 3y \):
\[
3y = 9 - 6x
\]

3. Divide by 3:
\[
y = 3 - 2x
\]

Now, the slope-intercept form of the first equation is:
\[
y = -2x + 3
\]

To create a second equation that has an infinite number of solutions, we can multiply the entire equation by a non-zero constant. A simple choice is to multiply by 2:

\[
2y = 2(-2x + 3)
\]
\[
2y = -4x + 6
\]

Now we rearrange it back to slope-intercept form:

\[
2y = -4x + 6
\]

Divide everything by 2:

\[
y = -2x + 3
\]

However, we can also express the same line differently. For example, if we multiply the original equation by \(\frac{1}{2}\):

\[
\frac{1}{2}(6x + 3y) = \frac{1}{2}(9)
\]

This results in:

\[
3x + \frac{3}{2}y = \frac{9}{2}
\]

Rearranging it to slope-intercept form leads to:

\[
\frac{3}{2}y = -3x + \frac{9}{2}
\]

Divide by \(\frac{3}{2}\):

\[
y = -2x + 3
\]

Thus, the simplest way to write a second equation that will also have an infinite number of solutions is simply to use the same line again, or a proportional modification. One example of a second equation in the slope-intercept form could simply be the same as above:

\[
y = -2x + 3
\]

So we can write:

The second equation could be \( y = -2x + 3 \).

Answer 5

To determine which points are solutions to the system of equations given by:

1. \( y = 2x + 2 \)
2. \( y = 2x - 4 \)

let’s check each point by substituting the \( x \) and \( y \) coordinates into both equations.

**Testing the point (0, −4):**

1. Substitute into the first equation \( y = 2x + 2 \):
\[
-4 = 2(0) + 2 \implies -4 = 2 \quad (\text{False})
\]

2. Substitute into the second equation \( y = 2x - 4 \):
\[
-4 = 2(0) - 4 \implies -4 = -4 \quad (\text{True})
\]

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

**Testing the point (0, 2):**

1. Substitute into the first equation \( y = 2x + 2 \):
\[
2 = 2(0) + 2 \implies 2 = 2 \quad (\text{True})
\]

2. Substitute into the second equation \( y = 2x - 4 \):
\[
2 = 2(0) - 4 \implies 2 = -4 \quad (\text{False})
\]

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

**Conclusion:**
Neither point \( (0, -4) \) nor point \( (0, 2) \) is a solution to the system of equations (i.e., neither point satisfies both equations simultaneously).

Thus, the correct response is:

**Neither of these points are solutions.**