Question

Which point is a solution to the simultaneous inequalities: y < 1/2 x and y > 1/3 x + 3

Answer 1

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.

1. **Consider the first inequality**:
\[
y < \frac{1}{2}x
\]
This represents a region below the line \(y = \frac{1}{2}x\).

2. **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 \((x, y)\):
1. 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.