Question

which statements are true about the graph of $y \leq 3x + 1$ and $y \geq -x + 2$? check all that apply.
\square the slope of one boundary line is 2.
\square both boundary lines are solid.
\square a solution to the system is (1, 3).
\square both inequalities are shaded below the boundary lines.
\square the boundary lines intersect.

Explanation:

1. Analyze the slope of boundary lines: The equations are \(y = 3x + 1\) (slope \(3\)) and \(y=-x + 2\) (slope \(-1\)), so the first statement is false.
2. Check line type: Inequalities with \(\leq\) and \(\geq\) use solid lines, so both boundary lines are solid (second statement true).
3. Test \((1, 3)\): For \(y\leq3x + 1\), \(3\leq3(1)+1=4\) (true); for \(y\geq -x + 2\), \(3\geq -1 + 2 = 1\) (true), so \((1, 3)\) is a solution (third statement true).
4. Shading direction: \(y\leq3x + 1\) shades below, \(y\geq -x + 2\) shades above, so fourth statement false.
5. Intersection: Two non - parallel lines (\(3

eq - 1\)) intersect, so fifth statement true.

Answer:

B. Both boundary lines are solid.
C. A solution to the system is (1, 3).
E. The boundary lines intersect.