Question

related questions: lesson 5: inequalities

click on the area that would be within the solution set for this system of inequalities.

q. \\(\

$$\begin{cases} y \\ge 3x + 1 \\\\ y \\ge -x - 3 \\end{cases}$$

\\)

Explanation:

Identify the boundary lines

Using the Boundary Line Analysis knowledge point

$$
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$$

Determine the shaded region for each inequality

Using the Graphing Linear Inequalities knowledge point

$$
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$$

Find the intersection of the shaded regions

To find the common solution set, we look for the region that is simultaneously above both boundary lines.

• The line \(y = 3x + 1\) goes upward steeply from left to right. The region "above" this line is to its left/upper-left.
• The line \(y = -x - 3\) goes downward from left to right. The region "above" this line is to its right/upper-right.
• The intersection of these two half-planes is the upper V-shaped region bounded by the two lines. This region contains the positive y-axis (for example, the point \((0, 2)\) satisfies both: \(2 \ge 1\) and \(2 \ge -3\)).

Identify a representative point in the solution set

Let's test a point in the upper region, such as \((0, 0)\):

1. For \(y \ge 3x + 1\): \(0 \ge 3(0) + 1 \implies 0 \ge 1\) (False). Thus, \((0,0)\) is not in the solution set.

Let's test \((-1, 2)\):

1. For \(y \ge 3x + 1\): \(2 \ge 3(-1) + 1 \implies 2 \ge -2\) (True).
2. For \(y \ge -x - 3\): \(2 \ge -(-1) - 3 \implies 2 \ge -2\) (True).

The region containing \((-1, 2)\) (the top region between the two intersecting lines) is the solution set.

Answer:

The solution set for this system of inequalities is the upper region bounded by both lines (the region containing the upper y-axis, above both boundary lines).

To select the correct area on the screen, click anywhere in the region above both intersecting lines (for example, near the point \((-1, 2)\) or \((0, 3)\)).