Question

Which point on the graph is a solution to the simultaneous inequalities y≤ 1/3x +2 and y< 2/3x +4

Answer 1

To find a point that satisfies the simultaneous inequalities \( y \leq \frac{1}{3}x + 2 \) and \( y < \frac{2}{3}x + 4 \), we will first analyze each inequality.

1. **Graph the first inequality:**
- The line \( y = \frac{1}{3}x + 2 \) has a slope of \( \frac{1}{3} \) and a y-intercept of 2.
- The region satisfies \( y \leq \frac{1}{3}x + 2 \), which means we are interested in the area below (and including) this line.

2. **Graph the second inequality:**
- The line \( y = \frac{2}{3}x + 4 \) has a slope of \( \frac{2}{3} \) and a y-intercept of 4.
- The region satisfies \( y < \frac{2}{3}x + 4 \), which means we are interested in the area below this line but not including it.

Next, we need to find a point that lies in the overlapping region defined by these two inequalities.

Let's test a few simple points:

- **Point (0, 1)**:
- For the first inequality:
\[ 1 \leq \frac{1}{3}(0) + 2 \Rightarrow 1 \leq 2 \quad \text{(True)} \]
- For the second inequality:
\[ 1 < \frac{2}{3}(0) + 4 \Rightarrow 1 < 4 \quad \text{(True)} \]
- So, (0, 1) is a solution.

- **Point (0, 5)**:
- For the first inequality:
\[ 5 \leq \frac{1}{3}(0) + 2 \Rightarrow 5 \leq 2 \quad \text{(False)} \]
- For the second inequality:
\[ 5 < \frac{2}{3}(0) + 4 \Rightarrow 5 < 4 \quad \text{(False)} \]
- So, (0, 5) is not a solution.

- **Point (-3, 1)**:
- For the first inequality:
\[ 1 \leq \frac{1}{3}(-3) + 2 \Rightarrow 1 \leq -1 + 2 \Rightarrow 1 \leq 1 \quad \text{(True)} \]
- For the second inequality:
\[ 1 < \frac{2}{3}(-3) + 4 \Rightarrow 1 < -2 + 4 \Rightarrow 1 < 2 \quad \text{(True)} \]
- So, (-3, 1) is also a solution.

In summary, both points (0, 1) and (-3, 1) satisfy the given simultaneous inequalities \( y \leq \frac{1}{3}x + 2 \) and \( y < \frac{2}{3}x + 4 \).

You can choose either of them: **(0, 1)** is an example of a solution.