Question

which statements are true about the graph of the system of linear inequalities? select two options.
y > 3x - 4
y ≤ (1/2)x + 1

• the graph of y > 3x - 4 has shading above a dashed line.
• the graph of y ≤ (1/2)x + 1 has shading below a dashed line.
• the graphs of the inequalities will intersect.
• there are no solutions to the system.
• the graphs of the two inequalities intersect the y - axis at (0, 1) and (0, 4).

Explanation:

1. For \( y > 3x - 4 \): The inequality is "greater than", so the line \( y = 3x - 4 \) is dashed (since the inequality is strict, \( > \) not \( \geq \)) and shading is above the line (because \( y \) values greater than the line's \( y \)-values are above it). So this statement is true.
2. For \( y \leq \frac{1}{2}x + 1 \): The inequality is "less than or equal to", so the line \( y=\frac{1}{2}x + 1 \) is solid (because of the "or equal to" part, \( \leq \) not \( < \)), and shading is below the line. The statement says "dashed line", which is incorrect.
3. The two lines \( y = 3x - 4 \) (slope 3, y - intercept - 4) and \( y=\frac{1}{2}x + 1 \) (slope \( \frac{1}{2} \), y - intercept 1) have different slopes, so they are not parallel and will intersect. So the statement "The graphs of the inequalities will intersect" is true.
4. The system has solutions because the regions defined by the two inequalities (one above a dashed line, one below a solid line) will overlap where the two regions intersect (since the lines intersect). So "There are no solutions to the system" is false.
5. For \( y > 3x - 4 \), when \( x = 0 \), \( y > - 4 \), so it intersects the y - axis at \( (0,-4) \) (the line \( y = 3x - 4 \) intersects the y - axis at \( (0,-4) \)), not \( (0,4) \). For \( y\leq\frac{1}{2}x + 1 \), when \( x = 0 \), \( y\leq1 \), so it intersects the y - axis at \( (0,1) \). So the statement about the y - intercepts is false.

Answer:

A. The graph of \( y > 3x - 4 \) has shading above a dashed line.
C. The graphs of the inequalities will intersect.