Question
The graph for the equation y = negative 2 x + 1 is shown below.
On a coordinate plane, a line with negative slope goes through (0, 1) and (1, negative 1).
If another equation is graphed so that the system has no solution, which equation could that be?
A. y = negative 2 (x minus one-half)
B. y = negative one-half (4 x + 2)
C. y = negative x + 1
D. y = negative one-half x + 2
On a coordinate plane, a line with negative slope goes through (0, 1) and (1, negative 1).
If another equation is graphed so that the system has no solution, which equation could that be?
A. y = negative 2 (x minus one-half)
B. y = negative one-half (4 x + 2)
C. y = negative x + 1
D. y = negative one-half x + 2
Answers
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Answered by
GPT-4o mini
To determine which equation would present a system with no solution when graphed alongside the equation \( y = -2x + 1 \), we need to identify which equation represents a line parallel to the original line. Remember that parallel lines have the same slope but different y-intercepts.
The given equation is \( y = -2x + 1 \), which has a slope of \(-2\). We need to find another equation with the same slope of \(-2\) but a different y-intercept.
Now let's evaluate the options:
A. \( y = -2(x - \frac{1}{2}) = -2x + 1 \) (parallel but same line, has a valid intercept at (1/2))
B. \( y = -\frac{1}{2}(4x + 2) = -2x - 1 \) (this line has a different slope of \(-2\) but different y-intercept)
C. \( y = -x + 1 \) (this line has a slope of \(-1\), which is not parallel)
D. \( y = -\frac{1}{2}x + 2 \) (this line has a slope of \(-\frac{1}{2}\), which is not parallel)
Among these:
- The only option with the same slope of \(-2\) and a different intercept is found in option **A**, but it turns out to be the same line.
Thus, the correct choice that creates a system with no solution must have a slope of \(-2\) and a different intercept.
So:
**Answer: A. y = -2(x - 1/2)** is invalid because it is the same line. No other options create parallel lines with a different slope.
Hence, technically no options represent a parallel line with the required condition. But upon clarification:
The original line was confirmed parallelism with sufficient scrutiny, confirming the only viable option represents variable intercept conditions thus it retains solute in practical confirmation to intersect in system.
The given equation is \( y = -2x + 1 \), which has a slope of \(-2\). We need to find another equation with the same slope of \(-2\) but a different y-intercept.
Now let's evaluate the options:
A. \( y = -2(x - \frac{1}{2}) = -2x + 1 \) (parallel but same line, has a valid intercept at (1/2))
B. \( y = -\frac{1}{2}(4x + 2) = -2x - 1 \) (this line has a different slope of \(-2\) but different y-intercept)
C. \( y = -x + 1 \) (this line has a slope of \(-1\), which is not parallel)
D. \( y = -\frac{1}{2}x + 2 \) (this line has a slope of \(-\frac{1}{2}\), which is not parallel)
Among these:
- The only option with the same slope of \(-2\) and a different intercept is found in option **A**, but it turns out to be the same line.
Thus, the correct choice that creates a system with no solution must have a slope of \(-2\) and a different intercept.
So:
**Answer: A. y = -2(x - 1/2)** is invalid because it is the same line. No other options create parallel lines with a different slope.
Hence, technically no options represent a parallel line with the required condition. But upon clarification:
The original line was confirmed parallelism with sufficient scrutiny, confirming the only viable option represents variable intercept conditions thus it retains solute in practical confirmation to intersect in system.
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