Question

Which point is the solution for y > 4x - 2 and y ≥ -1/3x + 2

Answer 1

To find a point that satisfies both inequalities:

1. **Inequality 1:** \(y > 4x - 2\)
2. **Inequality 2:** \(y \geq -\frac{1}{3}x + 2\)

Let's solve by substituting points into both inequalities.

**Choosing a point:**

We'll start with \(x = 0\):

1. For \(x = 0\) in the first inequality:
\[
y > 4(0) - 2 \implies y > -2
\]

2. For \(x = 0\) in the second inequality:
\[
y \geq -\frac{1}{3}(0) + 2 \implies y \geq 2
\]

Now, we need to check for a suitable \(y\) value that satisfies both conditions:
- From the first inequality: \(y > -2\)
- From the second inequality: \(y \geq 2\)

Since \(y\) must be greater than \(-2\) and at the same time \(y\) must be greater than or equal to \(2\), we can choose \(y = 2\) or any value greater than \(2\).

**Example of a valid point:**
Let's choose \(y = 3\), then the point \((0, 3)\) satisfies both inequalities:

1. **First Inequality:**
\[
3 > 4(0) - 2 \implies 3 > -2 \quad \text{(True)}
\]

2. **Second Inequality:**
\[
3 \geq -\frac{1}{3}(0) + 2 \implies 3 \geq 2 \quad \text{(True)}
\]

Thus, the point \((0, 3)\) is a solution for the inequalities \(y > 4x - 2\) and \(y \geq -\frac{1}{3}x + 2\).

Any point of the form \((x, y)\) where \(y > 2\) would also work, for example, \((1, 5)\) or \((2, 4)\) would also be valid solutions.