Question

Systems of linear inequalities solutions
Determine if each ordered pair is a solution of the system of linear inequalities.
#2 y<x-5 & y≤-2x+1
(0,-10) (8,1) (5,0) (2,5) (1,-6)

Answer 1

To determine if each ordered pair is a solution to the system of linear inequalities given by:

1. \( y < x - 5 \)
2. \( y \leq -2x + 1 \)

we will test each ordered pair \((x, y)\) to see if it satisfies both inequalities.

### Testing Each Ordered Pair

1. **For (0, -10)**:
- First inequality: \( -10 < 0 - 5 \) → \( -10 < -5 \) (True)
- Second inequality: \( -10 \leq -2(0) + 1 \) → \( -10 \leq 1 \) (True)
- Result: **(0, -10) is a solution.**

2. **For (8, 1)**:
- First inequality: \( 1 < 8 - 5 \) → \( 1 < 3 \) (True)
- Second inequality: \( 1 \leq -2(8) + 1 \) → \( 1 \leq -16 + 1 \) → \( 1 \leq -15 \) (False)
- Result: **(8, 1) is NOT a solution.**

3. **For (5, 0)**:
- First inequality: \( 0 < 5 - 5 \) → \( 0 < 0 \) (False)
- Second inequality (not needed since first is false, but let's check): \( 0 \leq -2(5) + 1 \) → \( 0 \leq -10 + 1 \) → \( 0 \leq -9 \) (False)
- Result: **(5, 0) is NOT a solution.**

4. **For (2, 5)**:
- First inequality: \( 5 < 2 - 5 \) → \( 5 < -3 \) (False)
- Second inequality (not needed since first is false, but let's check): \( 5 \leq -2(2) + 1 \) → \( 5 \leq -4 + 1 \) → \( 5 \leq -3 \) (False)
- Result: **(2, 5) is NOT a solution.**

5. **For (1, -6)**:
- First inequality: \( -6 < 1 - 5 \) → \( -6 < -4 \) (True)
- Second inequality: \( -6 \leq -2(1) + 1 \) → \( -6 \leq -2 + 1 \) → \( -6 \leq -1 \) (False)
- Result: **(1, -6) is NOT a solution.**

### Summary of Results
- (0, -10) **is a solution.**
- (8, 1) **is NOT a solution.**
- (5, 0) **is NOT a solution.**
- (2, 5) **is NOT a solution.**
- (1, -6) **is NOT a solution.**