Question

pretest: special linear relationships
which solution is valid within the context of the situation?
a. (-10,60)
b. (65,17.5)
c. (72,24)
d. (120,-6)

Explanation:

Step1: Find the equation of the line

The line passes through \((0, 50)\) and \((100, 0)\). The slope \(m=\frac{0 - 50}{100 - 0}=-\frac{1}{2}\). Using the slope - intercept form \(y=mx + b\) with \(b = 50\), the equation is \(y=-\frac{1}{2}x + 50\).

Step2: Check each option

- Option A: Substitute \(x=- 10\) into \(y =-\frac{1}{2}x+50\), we get \(y=-\frac{1}{2}\times(-10)+50 = 5 + 50=55 eq60\). Also, the point \((-10,60)\) is not in the blue - shaded region (the region above the line).
- Option B: Substitute \(x = 65\) into \(y=-\frac{1}{2}x + 50\), we get \(y=-\frac{1}{2}\times65 + 50=-32.5 + 50 = 17.5\). Now, check the region: The blue region is above the line. For \(x = 65\), the line has \(y = 17.5\), and \((65,17.5)\) is on the line? Wait, no, wait the blue region: Let's see the graph, the blue region is above the line. Wait, when \(x = 0\), the line is at \(y = 50\), and the blue region is above (since at \(x=-50\), the \(y\) - value of the line is \(y=-\frac{1}{2}\times(-50)+50 = 25 + 50 = 75\), and the blue region is above that). Wait, maybe I made a mistake in the region. Wait the line goes from \((-50,75)\) to \((200,-50)\) (by extending the line \(y =-\frac{1}{2}x + 50\)). The blue region is above the line. Let's check the \(y\) - value for each point relative to the line:
- For point A \((-10,60)\): The line at \(x=-10\) is \(y = 55\). \(60>55\), but let's check the \(x\) - range. The \(x\) - axis in the graph: the blue region seems to have \(x\) from \(-50\) to \(200\), but the point \((-10,60)\): Wait, when \(x=-10\), the line is \(y = 55\), and the blue region is above the line. But let's check the other points.
- For point B \((65,17.5)\): The line at \(x = 65\) is \(y=17.5\). The blue region is above the line, so \(y\) should be greater than \(17.5\) for the blue region. But \(17.5\) is equal to the line's \(y\) - value at \(x = 65\), so it's on the line, not in the blue region.
- For point C \((72,24)\): Substitute \(x = 72\) into the line equation \(y=-\frac{1}{2}\times72+50=-36 + 50 = 14\). Since \(24>14\), and we check the region: the blue region is above the line. So \(y = 24>14\) (the line's \(y\) - value at \(x = 72\)), and also check the \(x\) - range (\(72\) is between \(-50\) and \(200\)).
- For point D \((120,-6)\): The line at \(x = 120\) is \(y=-\frac{1}{2}\times120 + 50=-60 + 50=-10\). \(-6>-10\), but the blue region at \(x = 120\): the line is at \(y=-10\), and the blue region is above, but \(-6>-10\), but let's check the \(y\) - range. The blue region has \(y\) values above a certain level. Wait, maybe the correct way is to check if the point is in the blue - shaded area. The blue - shaded area is above the line. Let's re - express the inequality. The line is \(y=-\frac{1}{2}x + 50\), and the blue region satisfies \(y\geq-\frac{1}{2}x + 50\) (assuming the boundary is included).
- For A: \(60\geq-\frac{1}{2}\times(-10)+50=5 + 50 = 55\), \(60\geq55\) is true, but let's check the graph's \(x\) - limit. The left - most \(x\) in the graph is \(-50\), and \(-10\) is within \(-50\) to \(200\). But wait, when we look at the graph, the blue region at \(x=-50\) has a \(y\) - value of the line as \(y = 75\), and the blue region is above. But the point \((-10,60)\): the \(y\) - value of the line at \(x=-10\) is \(55\), and \(60>55\), but maybe the problem is about the context (maybe non - negative \(x\) and \(y\)? No, the graph has negative \(x\) and \(y\)). Wait, maybe I made a mistake in the equation. Let's re - calculate the equation of the line. The line passes through \((100,0)\) and \((0,50)\). So the slope \(m=\frac{0 - 50}{100 - 0}=-\frac{1}{2}\), so \(y=-\frac{1}{2}x + 50\) is correct.
- For point C \((72,24)\): The line at \(x = 72\) is \(y=-\frac{1}{2}\times72+50=-36 + 50 = 14\). \(24>14\), so the point \((72,24)\) is above the line (in the blue region).
- For point D \((120,-6)\): The line at \(x = 120\) is \(y=-10\). \(-6>-10\), but the \(y\) - value of the point is \(-6\), and the blue region at \(x = 120\): the line is at \(y=-10\), and the blue region is above, but \(-6\) is greater than \(-10\), but let's check the \(y\) - range of the blue region. The blue region has \(y\) values from, say, when \(x=-50\), \(y = 75\) down to \(y=-50\) at \(x = 200\), but the blue region is above the line. Wait, maybe the context is about non - negative quantities? No, the graph has negative axes. Wait, let's check the \(y\) - value of each point:
- Point A: \(y = 60\) (positive), point B: \(y = 17.5\) (positive), point C: \(y = 24\) (positive), point D: \(y=-6\) (negative). The blue region in the graph has positive \(y\) - values in most of it. The line at \(x = 120\) has \(y=-10\), and the blue region at \(x = 120\) is above the line, but \(y=-6\) is negative. The blue region above the line when \(x\) is between \(-50\) and \(100\) has positive \(y\) - values, and when \(x>100\), the line goes negative, and the blue region above the line also goes negative, but let's check the equation again.
Wait, maybe the inequality is \(y\geq-\frac{1}{2}x + 50\). Let's check each point:
- A: \(x=-10\), \(y = 60\). \(-\frac{1}{2}\times(-10)+50=5 + 50 = 55\). \(60\geq55\): true. But let's check the graph's \(x\) - limit. The left - most \(x\) in the graph is \(-50\), and \(-10\) is within \(-50\) to \(200\). But the \(y\) - value of the line at \(x=-10\) is \(55\), and the point is \((-10,60)\), which is above the line. But maybe the problem is that the context requires \(x\) and \(y\) to be such that the point is in the blue - shaded area. Wait, the blue - shaded area: when \(x = 0\), the line is at \(y = 50\), and the blue area is above (since at \(x=-50\), the line is at \(y = 75\), and the blue area is above that). At \(x = 50\), the line is at \(y=-\frac{1}{2}\times50 + 50=-25 + 50 = 25\), and the blue area is above (so \(y>25\) at \(x = 50\)).
- B: \(x = 65\), line \(y = 17.5\). The blue area at \(x = 65\) should have \(y>17.5\), but the point is \((65,17.5)\) which is on the line, not in the blue area (the blue area is above the line, so \(y>17.5\) at \(x = 65\)).
- C: \(x = 72\), line \(y=-\frac{1}{2}\times72 + 50=-36 + 50 = 14\). The blue area at \(x = 72\) has \(y>14\), and \(24>14\), so this point is in the blue area.
- D: \(x = 120\), line \(y=-10\). The blue area at \(x = 120\) has \(y>-10\), but \(y=-6>-10\), but the \(y\) - value is negative, and the blue area in the graph seems to have positive \(y\) - values for \(x\) between \(-50\) and \(100\), and negative \(y\) - values for \(x>100\), but the point \((120,-6)\) is in the blue area? Wait, no, when \(x = 120\), the line is at \(y=-10\), and the blue area is above the line, so \(y>-10\) at \(x = 120\), so \((120,-6)\) is in the blue area? But let's check the equation again. Wait, maybe the original line is \(y=-\frac{1}{2}x + 50\), and the blue region is \(y\geq-\frac{1}{2}x + 50\). Let's check the \(y\) - intercept and slope again. The line passes through \((0,50)\) and \((100,0)\), so slope is \(-\frac{1}{2}\), equation is correct.
Now, let's check the \(y\) - value of each point:
- A: \(y = 60\), line at \(x=-10\) is \(y = 55\), \(60>55\): in the region? But the graph's blue region at \(x=-10\) is above the line, but maybe the \(x\) - range is from \(-50\) to \(200\), and \(y\) - range from, say, \(-50\) to \(200\). But let's check the answer options. The most appropriate is C. Wait, maybe I made a mistake in the region. Let's re - plot the line: when \(x = 0\), \(y = 50\); \(x = 100\), \(y = 0\); \(x = 200\), \(y=-50\); \(x=-50\), \(y = 75\). The blue region is above the line. So for a point to be in the blue region, \(y\geq-\frac{1}{2}x + 50\).
- For A: \(60\geq-\frac{1}{2}\times(-10)+50=55\): true. But let's check the \(x\) - value: \(x=-10\) is valid, but the \(y\) - value of the point is \(60\), and the line at \(x=-10\) is \(55\), so it's in the region. But why is C the answer? Wait, maybe the context is about non - negative \(x\) and \(y\). If \(x\) and \(y\) are non - negative, then \(x\geq0\) and \(y\geq0\).
- A: \(x=-10<0\): invalid (if \(x\) is non - negative in context).
- B: \(x = 65\geq0\), \(y = 17.5\geq0\), but on the line.
- C: \(x = 72\geq0\), \(y = 24\geq0\), and above the line.
- D: \(y=-6<0\): invalid (if \(y\) is non - negative in context).
Ah! That's the key. In real - world linear relationships (like cost - profit, etc.), \(x\) and \(y\) are often non - negative. So:
- Option A: \(x=-10\) (negative \(x\)): invalid.
- Option B: The point \((65,17.5)\) is on the line (\(y=-\frac{1}{2}\times65 + 50=-32.5 + 50 = 17.5\)), so it's on the boundary, not in the interior of the blue region (the blue region is above the line, so the boundary is not included as a "solution" in the context, maybe the solution is in the interior).
- Option C: \(x = 72\geq0\), \(y = 24\geq0\), and \(y=24>-\frac{1}{2}\times72 + 50=14\), so it's in the blue region (above the line) and has non - negative \(x\) and \(y\).
- Option D: \(y=-6<0\): invalid.

Answer:

C. \((72,24)\)