Question 19 Pause The folowing system of equations has no solution. \[ \left\{\begin{array}{l} 2 x+y=1 \\ y=-2 x-3 \end{array}\right. \] When the fines are graphed on the same coordinate plane, they are A. perpendicular lines. B. intersecting lines. C. non-linear lines. D. parallel lines.

To determine the nature of the lines represented by the given system of equations, we need to analyze the equations: 1. The first equation is \(2x + y = 1\). 2. The second equation is \(y = -2x - 3\). ### Step 1: Rewrite the first equation in slope-intercept form We can rearrange the first equation to find its slope: \[ y = -2x + 1 \] ### Step 2: Identify the slopes Now we have both equations in slope-intercept form: 1. From \(y = -2x + 1\), the slope \(m_1 = -2\). 2. From \(y = -2x - 3\), the slope \(m_2 = -2\). ### Step 3: Compare the slopes Since both lines have the same slope (\(m_1 = m_2 = -2\)), they are parallel lines. ### Conclusion Since the problem states that the system of equations has no solution, this confirms that the lines do not intersect, which is characteristic of parallel lines. Thus, the correct answer is: **D. parallel lines.**