To find a point that satisfies both inequalities \(y < \frac{1}{2}x\) and \(y > \frac{1}{3}x + 3\), we can analyze the two inequalities and find the region they describe.
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Consider the first inequality: \[ y < \frac{1}{2}x \] This represents a region below the line \(y = \frac{1}{2}x\).
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Consider the second inequality: \[ y > \frac{1}{3}x + 3 \] This represents a region above the line \(y = \frac{1}{3}x + 3\).
Next, we need to find their intersection to visualize where these regions would overlap.
Finding Intersection Points:
To find where \(y = \frac{1}{2}x\) intersects with \(y = \frac{1}{3}x + 3\), set the equations equal to each other: \[ \frac{1}{2}x = \frac{1}{3}x + 3 \]
Multiply each term by 6 (the least common multiple of 2 and 3) to eliminate fractions: \[ 3x = 2x + 18 \] \[ 3x - 2x = 18 \] \[ x = 18 \]
Now substitute \(x = 18\) back into either equation to find \(y\): Using \(y = \frac{1}{2}x\): \[ y = \frac{1}{2}(18) = 9 \]
So, the intersection point is \((18, 9)\).
Definition of Regions:
- The area below the line \(y = \frac{1}{2}x\) includes points where \(x < 18\), and \(y < 9\).
- The area above the line \(y = \frac{1}{3}x + 3\) includes points that are greater than the line.
Combine these areas, we need to find a point \((x, y)\):
- For example, let's take \(x = 0\):
- From \(y < \frac{1}{2}(0) = 0\), we have \(y < 0\).
- From \(y > \frac{1}{3}(0) + 3 = 3\), we have \(y > 3\).
Thus, there is no solution in this region (no overlapping).
Now test a different \(x\): Let's try \(x = 24\):
- For \(y < \frac{1}{2}(24) = 12\)
- For \(y > \frac{1}{3}(24) + 3 = 8 + 3 = 11\)
This gives us:
- Must satisfy: \(11 < y < 12\), hence \(y\) could be \(y = 11.5\).
So a point that satisfies both inequalities is: \[ (24, 11.5) \]
This would be a valid solution to the simultaneous inequalities. Therefore, \((24, 11.5)\) is a solution.