Question

which system of inequalities does the graph represent? which test point satisfies both of the inequalities in that system?
the graph represents the system of inequalities
the test point

satisfies both of the ine

2x + 3y is greater than or equal to 4 and x + 2y is greater than or equal to 3
2x + 3y is less than or equal to 4 and x + 2y is less than or equal to 3
2x + 3y is greater than or equal to 4 and x + 2y is less than or equal to 3
2x + 3y is less than or equal to 4 and x + 2y is greater than or equal to 3
2x + 3y is less than or equal to 4 and 2x + 2y is less than or equal to 3

Explanation:

Step1: Analyze the first line (2x + 3y)

The blue line (first inequality) has a y-intercept. Let's find its equation. From the graph, when x=0, y-intercept for blue line: let's assume the line is \(2x + 3y = 4\) (since when x=0, \(3y = 4\) → \(y=\frac{4}{3}\), but wait, maybe better to check the shaded region. The shaded region for blue is below? Wait, no, the test point and the inequalities. Wait, the first inequality: let's rewrite \(2x + 3y \leq 4\) (since the line is solid, so equal included, and shaded below? Wait, no, the graph's shaded area: looking at the two lines, blue and red. Let's find the equations.

For the blue line: Let's take two points. When x=0, y-intercept: from the graph, blue line at x=0, y is around \(\frac{4}{3}\)? Wait, maybe the first inequality is \(2x + 3y \leq 4\) (solid line, shaded below). For the red line: let's find its equation. When x=0, y-intercept: red line at x=0, y=1.5? Wait, maybe \(x + 2y \leq 3\)? Wait, no, the options have \(2x + 3y \leq 4\) and \(2x + 2y \leq 3\)? No, the options are:

Wait the options are:

1. \(2x + 3y \geq 4\) and \(x + 2y \geq 3\)

2. \(2x + 3y \leq 4\) and \(x + 2y \leq 3\)

3. \(2x + 3y \geq 4\) and \(x + 2y \leq 3\)

4. \(2x + 3y \leq 4\) and \(x + 2y \geq 3\)

Wait, the graph: the two lines, blue and red. The shaded region is where both inequalities are satisfied. Let's check the test point. Let's take a test point, say (0,0). Plug into \(2x + 3y\): 0 + 0 = 0 ≤ 4? Yes. For \(x + 2y\): 0 + 0 = 0 ≤ 3? No, wait, no. Wait, maybe the red line is \(x + 2y \geq 3\)? Wait, no. Wait the options: the fourth option is \(2x + 3y \leq 4\) and \(x + 2y \geq 3\). Let's check (0,0) in \(x + 2y \geq 3\): 0 ≥ 3? No. So (0,0) is not in the shaded region. The shaded region is below blue and above red? Wait, the graph's shaded area: looking at the two lines, blue (solid) and red (solid). The intersection point: solve \(2x + 3y = 4\) and \(x + 2y = 3\). Multiply second equation by 2: \(2x + 4y = 6\), subtract first: \(y = 2\), then x = 3 - 2y = 3 -4 = -1. So intersection at (-1, 2). Now, the shaded region: between the two lines? Wait, no, the test point. Let's take a point in the shaded region, say (-2, 0). Plug into \(2x + 3y\): 2*(-2) + 0 = -4 ≤ 4? Yes. Plug into \(x + 2y\): -2 + 0 = -2 ≥ 3? No. Wait, maybe I got the inequalities reversed. Wait, the first inequality: \(2x + 3y \leq 4\) (solid, shaded below), second inequality: \(x + 2y \geq 3\) (solid, shaded above). So the overlapping region is where \(2x + 3y \leq 4\) and \(x + 2y \geq 3\). Let's check the test point. Let's take (-1, 2): plug into \(2x + 3y\): 2*(-1) + 3*2 = -2 +6 =4 ≤4? Yes. Plug into \(x + 2y\): -1 +4 =3 ≥3? Yes. So that point is on both lines. Now, the options: the fourth option is "2x + 3y is less than or equal to 4 and x + 2y is greater than or equal to 3" (the fourth option in the list: "2x + 3y is less than or equal to 4 and x + 2y is greater than or equal to 3").

Now, test point: let's take (-1, 2) which is on both lines. Or another point, say (0, 2): plug into \(2x + 3y\): 0 +6=6 ≤4? No. Wait, maybe ( -2, 1): \(2*(-2) +3*1= -4 +3= -1 ≤4\) (yes), \(x + 2y= -2 +2=0 ≥3\)? No. Wait, maybe the correct system is \(2x + 3y \leq 4\) and \(x + 2y \geq 3\), which is the fourth option (the one with "2x + 3y is less than or equal to 4 and x + 2y is greater than or equal to 3").

For the test point: let's take the intersection point (-1, 2) as a test point (since it's on both lines). Plug into both: \(2*(-1) +3*2= -2 +6=4 ≤4\) (satisfies first), \( -1 +2*2= -1 +4=3 ≥3\) (satisfies second). So that's a test point.

Step2: Confirm the system

The two lines are solid (so equal included), and the shaded region is where \(2x + 3y \leq 4\) (shaded below blue line) and \(x + 2y \geq 3\) (shaded above red line). So the system is \(2x + 3y \leq 4\) and \(x + 2y \geq 3\), which matches the fourth option in the list (the one with "2x + 3y is less than or equal to 4 and x + 2y is greater than or equal to 3").

Answer:

The system of inequalities is: 2x + 3y is less than or equal to 4 and x + 2y is greater than or equal to 3 (the fourth option in the list). A test point that satisfies both is (-1, 2) (or the intersection point of the two lines).