Question

Which statements are true about the graph of y ≤ 3x + 1 and y ≥ –x + 2? Check all that apply. 1.The slope of one boundary line is 2.
2.Both boundary lines are solid.
3.A solution to the system is (1, 3).
4.Both inequalities are shaded below the boundary lines.
5.The boundary lines intersect.

Answer 1

Answer:

2.Both boundary lines are solid.

3.A solution to the system is (1, 3)

5.The boundary lines intersect.

Step-by-step explanation:

we have

$y\leq 3x+1$ ----> inequality A

The solution of the inequality A is the shaded area below the solid line $y=3x+1$

The slope of the solid line is 3

The point (1,3) is  a solution of inequality A (lie in the shaded area of the solution set)

$y\geq -x+2$ ----> inequality B

The solution of the inequality B is the shaded area above the solid line $y=-x+2$

The slope of the solid line is -1

The point (1,3) is a solution of inequality B (lie in the shaded area of the solution set)

The solution of the system of inequalities is the shaded area between the two solids lines

see the attached figure

Verify each statement

1.The slope of one boundary line is 2

The statement is False

2.Both boundary lines are solid.

The statement is True

3.A solution to the system is (1, 3)

The statement is True

4.Both inequalities are shaded below the boundary lines

The statement is False

5.The boundary lines intersect.

The statement is True

The intersection point is (0.25,1.75)

see the attached figure

Answer 2

Answer:

2.Both boundary lines are solid.

3.A solution to the system is (1, 3)

5.The boundary lines intersect.

Step-by-step explanation: