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.
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