Answer :
Certainly! Let's analyze each of the points \((1,0)\), \((-1,1)\), \((2,2)\), and \((0,3)\) to determine if they satisfy both inequalities:
1. \( y > -3x + 3 \)
2. \( y \geq 2x - 2 \)
### Checking Point \((1, 0)\)
For the inequalities:
1. \( 0 > -3(1) + 3 \)
- Calculate the right side: \( -3(1) + 3 = -3 + 3 = 0 \)
- The inequality becomes \( 0 > 0 \), which is false.
2. \( 0 \geq 2(1) - 2 \)
- Calculate the right side: \( 2(1) - 2 = 2 - 2 = 0 \)
- The inequality becomes \( 0 \geq 0 \), which is true.
Since the first inequality is false, the point \((1, 0)\) does not satisfy both inequalities.
### Checking Point \((-1, 1)\)
For the inequalities:
1. \( 1 > -3(-1) + 3 \)
- Calculate the right side: \( -3(-1) + 3 = 3 + 3 = 6 \)
- The inequality becomes \( 1 > 6 \), which is false.
2. \( 1 \geq 2(-1) - 2 \)
- Calculate the right side: \( 2(-1) - 2 = -2 - 2 = -4 \)
- The inequality becomes \( 1 \geq -4 \), which is true.
Since the first inequality is false, the point \((-1, 1)\) does not satisfy both inequalities.
### Checking Point \((2, 2)\)
For the inequalities:
1. \( 2 > -3(2) + 3 \)
- Calculate the right side: \( -3(2) + 3 = -6 + 3 = -3 \)
- The inequality becomes \( 2 > -3 \), which is true.
2. \( 2 \geq 2(2) - 2 \)
- Calculate the right side: \( 2(2) - 2 = 4 - 2 = 2 \)
- The inequality becomes \( 2 \geq 2 \), which is true.
Since both inequalities are true, the point \((2, 2)\) does satisfy both inequalities.
### Checking Point \((0, 3)\)
For the inequalities:
1. \( 3 > -3(0) + 3 \)
- Calculate the right side: \( -3(0) + 3 = 0 + 3 = 3 \)
- The inequality becomes \( 3 > 3 \), which is false.
2. \( 3 \geq 2(0) - 2 \)
- Calculate the right side: \( 2(0) - 2 = 0 - 2 = -2 \)
- The inequality becomes \( 3 \geq -2 \), which is true.
Since the first inequality is false, the point \((0, 3)\) does not satisfy both inequalities.
### Summary
Only the point \((2, 2)\) satisfies both inequalities. The results for each point are:
- \((1, 0)\): false
- \((-1, 1)\): false
- \((2, 2)\): true
- \((0, 3)\): false
So, the final results are:
[tex]\[ [False, False, True, False] \][/tex]
1. \( y > -3x + 3 \)
2. \( y \geq 2x - 2 \)
### Checking Point \((1, 0)\)
For the inequalities:
1. \( 0 > -3(1) + 3 \)
- Calculate the right side: \( -3(1) + 3 = -3 + 3 = 0 \)
- The inequality becomes \( 0 > 0 \), which is false.
2. \( 0 \geq 2(1) - 2 \)
- Calculate the right side: \( 2(1) - 2 = 2 - 2 = 0 \)
- The inequality becomes \( 0 \geq 0 \), which is true.
Since the first inequality is false, the point \((1, 0)\) does not satisfy both inequalities.
### Checking Point \((-1, 1)\)
For the inequalities:
1. \( 1 > -3(-1) + 3 \)
- Calculate the right side: \( -3(-1) + 3 = 3 + 3 = 6 \)
- The inequality becomes \( 1 > 6 \), which is false.
2. \( 1 \geq 2(-1) - 2 \)
- Calculate the right side: \( 2(-1) - 2 = -2 - 2 = -4 \)
- The inequality becomes \( 1 \geq -4 \), which is true.
Since the first inequality is false, the point \((-1, 1)\) does not satisfy both inequalities.
### Checking Point \((2, 2)\)
For the inequalities:
1. \( 2 > -3(2) + 3 \)
- Calculate the right side: \( -3(2) + 3 = -6 + 3 = -3 \)
- The inequality becomes \( 2 > -3 \), which is true.
2. \( 2 \geq 2(2) - 2 \)
- Calculate the right side: \( 2(2) - 2 = 4 - 2 = 2 \)
- The inequality becomes \( 2 \geq 2 \), which is true.
Since both inequalities are true, the point \((2, 2)\) does satisfy both inequalities.
### Checking Point \((0, 3)\)
For the inequalities:
1. \( 3 > -3(0) + 3 \)
- Calculate the right side: \( -3(0) + 3 = 0 + 3 = 3 \)
- The inequality becomes \( 3 > 3 \), which is false.
2. \( 3 \geq 2(0) - 2 \)
- Calculate the right side: \( 2(0) - 2 = 0 - 2 = -2 \)
- The inequality becomes \( 3 \geq -2 \), which is true.
Since the first inequality is false, the point \((0, 3)\) does not satisfy both inequalities.
### Summary
Only the point \((2, 2)\) satisfies both inequalities. The results for each point are:
- \((1, 0)\): false
- \((-1, 1)\): false
- \((2, 2)\): true
- \((0, 3)\): false
So, the final results are:
[tex]\[ [False, False, True, False] \][/tex]
