Question

which is true about the solution to the system of inequalities shown?
y ≥ 3x + 1
y ≤ 3x − 3
(image of a coordinate plane with two lines and shaded regions)
options:

• only values that satisfy y ≥ 3x + 1 are solutions.
• only values that satisfy y ≤ 3x − 3 are solutions.
• values that satisfy either y ≥ 3x + 1 or y ≤ 3x − 3 are solutions.
• there are no solutions.

Explanation:

Step1: Analyze the lines' slopes

The two inequalities are \( y \geq 3x + 1 \) and \( y \leq 3x - 3 \). The slope of both lines is \( 3 \), so they are parallel (same slope, different y - intercepts: \( 1 \) and \( - 3 \)).

Step2: Analyze the solution regions

The region for \( y \geq 3x + 1 \) is above the line \( y = 3x+1 \), and the region for \( y \leq 3x - 3 \) is below the line \( y = 3x - 3 \). Since the lines are parallel and the region above \( y = 3x + 1 \) and the region below \( y = 3x - 3 \) do not overlap (because \( 3x + 1>3x - 3 \) for all real \( x \)), there is no point that satisfies both inequalities simultaneously.

Answer:

There are no solutions.