Question
Which graph shows the solution to the system of linear equations?
y equals one half times x
x + 2y = −8
coordinate plane with one line that passes through the points 0 comma negative 4 and 2 comma negative 5 and another line that passes through the points 0 comma 0 and 2 comma 1
coordinate plane with one line that passes through the points 0 comma 2 and negative 3 comma 3 and another line that passes through the points 0 comma 0 and negative 3 comma negative 1
coordinate plane with one line that passes through the points 3 comma negative 3 and 0 comma negative 2 and another line that passes through the points 0 comma 0 and 3 comma 1
coordinate plane with one line that passes through the points 0 comma 4 and negative 1 comma 1 and another line that passes through the points 0 comma 0 and 1 comma 3
y equals one half times x
x + 2y = −8
coordinate plane with one line that passes through the points 0 comma negative 4 and 2 comma negative 5 and another line that passes through the points 0 comma 0 and 2 comma 1
coordinate plane with one line that passes through the points 0 comma 2 and negative 3 comma 3 and another line that passes through the points 0 comma 0 and negative 3 comma negative 1
coordinate plane with one line that passes through the points 3 comma negative 3 and 0 comma negative 2 and another line that passes through the points 0 comma 0 and 3 comma 1
coordinate plane with one line that passes through the points 0 comma 4 and negative 1 comma 1 and another line that passes through the points 0 comma 0 and 1 comma 3
Answers
GPT-4o mini
To solve for the graph that shows the system of linear equations given by:
1. \( y = \frac{1}{2}x \)
2. \( x + 2y = -8 \)
We can begin by rewriting the second equation in slope-intercept form (y = mx + b).
Starting with the second equation:
\[
x + 2y = -8
\]
Subtract \(x\) from both sides:
\[
2y = -x - 8
\]
Now, divide every term by 2 to solve for \(y\):
\[
y = -\frac{1}{2}x - 4
\]
Now we have both equations:
1. \( y = \frac{1}{2}x \) (a line with a positive slope that passes through the origin)
2. \( y = -\frac{1}{2}x - 4 \) (a line with a negative slope, crossing the y-axis at -4)
Next, we can identify a few key points on each line:
- For the first equation \( y = \frac{1}{2}x \):
- When \( x = 0 \), \( y = 0 \) (point: (0, 0))
- When \( x = 2 \), \( y = 1 \) (point: (2, 1))
- For the second equation \( y = -\frac{1}{2}x - 4 \):
- When \( x = 0 \), \( y = -4 \) (point: (0, -4))
- When \( x = -8 \), \( y = 0 \) (which means \( x = -8 \) gives: when \( y = 0 \), then \( x = -8 \))
Now, let’s check the provided graphs:
1. Graph 1 has lines passing through (0, -4) and (2, -5) (does not match either equation).
2. Graph 2 has lines passing through (0, 2) and (-3, 3) which do not align with our equations.
3. Graph 3 has lines passing through (3, -3) and (0, -2) (does not match).
4. Graph 4 has lines passing through (0, 4) and (-1, 1) which also do not align with our equations.
Since none of these options provided directly correspond to the equations we derived, it would appear that none of the provided graphs are correct based on the analysis of the equations. However, if you have graphical representation options to choose from, look for a graph where:
- One line goes through points like (0, 0) and (2, 1),
- The second line goes through points like (0, -4) and perhaps crosses the x-axis at -8.
If any graph matches these criteria closely, that would be the solution to the system of equations.
1. \( y = \frac{1}{2}x \)
2. \( x + 2y = -8 \)
We can begin by rewriting the second equation in slope-intercept form (y = mx + b).
Starting with the second equation:
\[
x + 2y = -8
\]
Subtract \(x\) from both sides:
\[
2y = -x - 8
\]
Now, divide every term by 2 to solve for \(y\):
\[
y = -\frac{1}{2}x - 4
\]
Now we have both equations:
1. \( y = \frac{1}{2}x \) (a line with a positive slope that passes through the origin)
2. \( y = -\frac{1}{2}x - 4 \) (a line with a negative slope, crossing the y-axis at -4)
Next, we can identify a few key points on each line:
- For the first equation \( y = \frac{1}{2}x \):
- When \( x = 0 \), \( y = 0 \) (point: (0, 0))
- When \( x = 2 \), \( y = 1 \) (point: (2, 1))
- For the second equation \( y = -\frac{1}{2}x - 4 \):
- When \( x = 0 \), \( y = -4 \) (point: (0, -4))
- When \( x = -8 \), \( y = 0 \) (which means \( x = -8 \) gives: when \( y = 0 \), then \( x = -8 \))
Now, let’s check the provided graphs:
1. Graph 1 has lines passing through (0, -4) and (2, -5) (does not match either equation).
2. Graph 2 has lines passing through (0, 2) and (-3, 3) which do not align with our equations.
3. Graph 3 has lines passing through (3, -3) and (0, -2) (does not match).
4. Graph 4 has lines passing through (0, 4) and (-1, 1) which also do not align with our equations.
Since none of these options provided directly correspond to the equations we derived, it would appear that none of the provided graphs are correct based on the analysis of the equations. However, if you have graphical representation options to choose from, look for a graph where:
- One line goes through points like (0, 0) and (2, 1),
- The second line goes through points like (0, -4) and perhaps crosses the x-axis at -8.
If any graph matches these criteria closely, that would be the solution to the system of equations.