Question

To determine which points are part of the solution set for the inequality \(y \geq 3x - 1\), we can substitute each point into the inequality and check if it holds true.
1. For the point \((1, 1)\):
\[
y \geq 3x - 1 \implies 1 \geq 3(1) - 1 \implies 1 \geq 3 - 1 \implies 1 \geq 2 \quad (\text{False})
\]

2. For the point \((-1, -5)\):
\[
y \geq 3x - 1 \implies -5 \geq 3(-1) - 1 \implies -5 \geq -3 - 1 \implies -5 \geq -4 \quad (\text{False})
\]

3. For the point \((2, 5)\):
\[
y \geq 3x - 1 \implies 5 \geq 3(2) - 1 \implies 5 \geq 6 - 1 \implies 5 \geq 5 \quad (\text{True})
\]

4. For the point \((-2.5, -9)\):
\[
y \geq 3x - 1 \implies -9 \geq 3(-2.5) - 1 \implies -9 \geq -7.5 - 1 \implies -9 \geq -8.5 \quad (\text{False})
\]

The only point that satisfies the inequality \(y \geq 3x - 1\) is \((2, 5)\). Therefore, the point \((2, 5)\) is a part of the solution set for the inequality.
make this like a 6th grader wrote it

Answer 1

Hey! Let’s see which points are part of the solution for the inequality \(y \geq 3x - 1\). To do this, we can plug in each point and check if it works.

1. **For the point \((1, 1)\)**:
\[
y \geq 3x - 1 \implies 1 \geq 3(1) - 1 \implies 1 \geq 3 - 1 \implies 1 \geq 2
\]
This is **False!** So \((1, 1)\) is not part of the solution.

2. **For the point \((-1, -5)\)**:
\[
y \geq 3x - 1 \implies -5 \geq 3(-1) - 1 \implies -5 \geq -3 - 1 \implies -5 \geq -4
\]
This is also **False!** So \((-1, -5)\) is not part of the solution.

3. **For the point \((2, 5)\)**:
\[
y \geq 3x - 1 \implies 5 \geq 3(2) - 1 \implies 5 \geq 6 - 1 \implies 5 \geq 5
\]
This one is **True!** So \((2, 5)\) is part of the solution.

4. **For the point \((-2.5, -9)\)**:
\[
y \geq 3x - 1 \implies -9 \geq 3(-2.5) - 1 \implies -9 \geq -7.5 - 1 \implies -9 \geq -8.5
\]
This is **False!** So \((-2.5, -9)\) is not part of the solution.

So, the only point that works for the inequality \(y \geq 3x - 1\) is \((2, 5)\)! That means \((2, 5)\) is in the solution set! 🎉