Question

the graph of a system of linear inequalities is shown
which of the followin points is a solution to the system?
a (9, 1)
b (3, 5)
c (4, 7)
d (7, 4)

Explanation:

Step1: Recall Solution Region

The solution to a system of linear inequalities is the region where all inequalities' solution regions overlap (the shaded area, including boundaries if solid lines). We check each point against this region.

Step2: Analyze Point (9,1)

Check if (9,1) lies in the shaded region. From the graph, the rightmost boundary of the shaded area (from the steep line) is at \( x = 7 \) (since the steep line ends at \( (7,0) \)). So \( x = 9 \) is to the right of \( x = 7 \), outside the shaded region. Eliminate (9,1).

Step3: Analyze Point (3,5)

Check (3,5). The shaded region spans \( x \) from 0 to 7 (approx, from the two lines) and \( y \) values within the two lines. At \( x = 3 \), \( y = 5 \) is within the shaded area (between the two lines, and \( x \leq 7 \)). Let's verify the other points to confirm.

Step4: Analyze Point (4,7)

Check (4,7). The upper line (the shallow one) at \( x = 4 \) has a \( y \)-value less than 7 (since it slopes downward from \( (0,8) \)). So \( y = 7 \) is above this line, outside the shaded region. Eliminate (4,7).

Step5: Analyze Point (7,4)

Check (7,4). The steep line at \( x = 7 \) has \( y = 0 \), but the shallow line at \( x = 7 \) has \( y \)-value: let's find its equation. The shallow line goes from \( (0,8) \) to, say, \( (10,2) \) (slope \( \frac{2 - 8}{10 - 0} = -\frac{3}{5} \), equation \( y = -\frac{3}{5}x + 8 \)). At \( x = 7 \), \( y = -\frac{21}{5} + 8 = -\frac{21}{5} + \frac{40}{5} = \frac{19}{5} = 3.8 \approx 4 \)? Wait, but the steep line at \( x = 7 \) is \( y = 0 \) (vertical line? No, the steep line is a line from, say, \( (7,9) \) to \( (7,0) \)? Wait, the graph shows a steep line going up to \( (7,9) \)? Wait, no, the original graph: the steep line has a vertex at \( (7,0) \) and goes up to \( (7,9) \)? Wait, maybe my initial analysis of the steep line was wrong. Wait, the shaded region is between the two lines: one is a shallow line from \( (0,8) \) to \( (10,2) \), and the other is a steep line from \( (7,0) \) up to \( (7,9) \)? No, the steep line is a vertical line? No, the steep line is a line with a steep slope, maybe from \( (7,0) \) to \( (6,9) \)? Wait, maybe I misread the graph. But the key is: the solution region is the intersection of the two inequalities. Let's re-express: the shaded area is bounded by two lines: one is a line with positive \( y \)-intercept (shallow, decreasing) and one is a line with a steep slope (maybe vertical? No, the steep line has a point at \( (7,0) \) and goes up. Wait, the correct approach is: a solution point must lie in the shaded area (where both inequalities are satisfied).

Wait, let's re-express the points:

- (3,5): \( x = 3 \), \( y = 5 \). The shaded area at \( x = 3 \) includes \( y \) between the two lines. The shallow line at \( x = 3 \): \( y = -\frac{3}{5}(3) + 8 = -\frac{9}{5} + 8 = \frac{31}{5} = 6.2 \). The steep line at \( x = 3 \): since the steep line is from \( (7,0) \) up, maybe it's a line with slope, say, from \( (7,0) \) to \( (0,9) \)? No, the graph's shaded area is between \( x = 0 \) to \( x = 7 \), and between the two lines. So (3,5): \( x = 3 \) is between 0 and 7, \( y = 5 \) is between the two lines (since the shallow line at \( x = 3 \) is \( y = -\frac{3}{5}(3) + 8 = 6.2 \), and the steep line at \( x = 3 \) is higher? Wait, maybe the steep line is the upper boundary? No, the shaded area is below the shallow line and above the steep line? Wait, the shaded area is gray, so let's assume the shaded region is where \( x \leq 7 \) (from the steep line ending at \( (7,0) \)) and below the shallow line (from \( (0,8) \) to \( (10,2) \)) and above the steep line (from \( (7,0) \) up). Wait, maybe I messed up the direction. Let's use the process of elimination again:

- (3,5): inside the shaded area (x between 0-7, y between the two lines).
- (9,1): x=9 >7 (steep line at x=7), outside.
- (4,7): y=7 is above the shallow line (at x=4, shallow line y= -3/5*4 +8= -12/5 +40/5=28/5=5.6 <7), so outside.
- (7,4): at x=7, the steep line is at y=0, but the shallow line at x=7 is y= -3/5*7 +8= -21/5 +40/5=19/5=3.8. So y=4 is above 3.8, but is x=7 allowed? The steep line at x=7: if it's a vertical line at x=7 (from (7,0) up), then x=7 is a boundary. But the shallow line at x=7 is y=3.8, so y=4 is above that, but is the shaded area above the steep line (y≥0) and below the shallow line (y≤ -3/5 x +8) and x≤7? Wait, maybe the shaded area is x from 0 to7, y from the steep line (y≥ something) to the shallow line (y≤ -3/5 x +8). Wait, the steep line: looking at the graph, the steep line goes from (7,0) up to (7,9) (vertical line x=7), and the shallow line from (0,8) to (10,2). So the shaded area is x ≤7 (left of x=7), and between the two lines (above the steep line? No, the steep line is at x=7, vertical, so left of x=7, and below the shallow line (y ≤ -3/5 x +8) and above the x-axis? Wait, the shaded area is gray, so let's check (3,5): x=3 ≤7, y=5 ≤ -3/5*3 +8=6.2, and y≥0 (since the steep line at x=3 is above y=0). So (3,5) is in the shaded area. (7,4): x=7 (boundary), y=4. The shallow line at x=7 is y=3.8, so y=4 is above that, so outside the shaded area (since shaded is below the shallow line). So (3,5) is the solution.

Answer:

B. (3, 5)