To see which of the points is a solution for the inequality \( y \leq \frac{2}{5}x + 1 \), we will plug in the \( x \) and \( y \) values for each point into the inequality.

Question

The inequality is:

\[
y \leq \frac{2}{5}x + 1
\]

Now let's check each point:

1. **Point (10, 6)**:
- Here, \( y = 6 \) and \( x = 10 \).
- We check:
\[
6 \leq \frac{2}{5}(10) + 1
\]
- Calculate the right side:
\[
\frac{2}{5}(10) = 4 \quad \text{so} \quad 4 + 1 = 5
\]
- Now we have:
\[
6 \leq 5 \quad \text{(This is False)}
\]

2. **Point (-10, 6)**:
- Here, \( y = 6 \) and \( x = -10 \).
- We check:
\[
6 \leq \frac{2}{5}(-10) + 1
\]
- Calculate the right side:
\[
\frac{2}{5}(-10) = -4 \quad \text{so} \quad -4 + 1 = -3
\]
- Now we have:
\[
6 \leq -3 \quad \text{(This is False)}
\]

3. **Point (10, 5.5)**:
- Here, \( y = 5.5 \) and \( x = 10 \).
- We check:
\[
5.5 \leq \frac{2}{5}(10) + 1
\]
- Calculate the right side:
\[
\frac{2}{5}(10) = 4 \quad \text{so} \quad 4 + 1 = 5
\]
- Now we have:
\[
5.5 \leq 5 \quad \text{(This is False)}
\]

4. **Point (10, 4)**:
- Here, \( y = 4 \) and \( x = 10 \).
- We check:
\[
4 \leq \frac{2}{5}(10) + 1
\]
- Calculate the right side:
\[
\frac{2}{5}(10) = 4 \quad \text{so} \quad 4 + 1 = 5
\]
- Now we have:
\[
4 \leq 5 \quad \text{(This is True)}
\]

So after checking all the points, the only point that works with the inequality \( y \leq \frac{2}{5}x + 1 \) is **(10, 4)**.

Therefore, **(10, 4)** is part of the solution set for the inequality!
make this like a 6th grader wrote this

Answer 1

To see which points are solutions for the inequality \( y \leq \frac{2}{5}x + 1 \), we need to plug in the \( x \) and \( y \) values for each point into the inequality.

The inequality we have is:

\[ y \leq \frac{2}{5}x + 1 \]

Now, let's check each point one by one:

1. Point (10, 6):

• For this point, \( y = 6 \) and \( x = 10 \).
• Let's check: \[ 6 \leq \frac{2}{5}(10) + 1 \]
• First, we calculate the right side: \[ \frac{2}{5}(10) = 4, \quad \text{so} \quad 4 + 1 = 5 \]
• Putting it together, we get: \[ 6 \leq 5 \quad \text{(This is False)} \]

2. Point (-10, 6):

• Here, \( y = 6 \) and \( x = -10 \).
• Checking again: \[ 6 \leq \frac{2}{5}(-10) + 1 \]
• Calculate the right side: \[ \frac{2}{5}(-10) = -4, \quad \text{so} \quad -4 + 1 = -3 \]
• Now we check: \[ 6 \leq -3 \quad \text{(This is False)} \]

3. Point (10, 5.5):

• Now we have \( y = 5.5 \) and \( x = 10 \).
• Let's see: \[ 5.5 \leq \frac{2}{5}(10) + 1 \]
• Right side: \[ \frac{2}{5}(10) = 4, \quad \text{so} \quad 4 + 1 = 5 \]
• Now check: \[ 5.5 \leq 5 \quad \text{(This is False)} \]

4. Point (10, 4):

• Lastly, for this point, \( y = 4 \) and \( x = 10 \).
• We check: \[ 4 \leq \frac{2}{5}(10) + 1 \]
• Calculate the right side: \[ \frac{2}{5}(10) = 4, \quad \text{so} \quad 4 + 1 = 5 \]
• Now: \[ 4 \leq 5 \quad \text{(This is True)} \]

So after checking all the points, the only one that works with the inequality \( y \leq \frac{2}{5}x + 1 \) is (10, 4).

That means, (10, 4) is part of the solution set for the inequality! 🎉