In which of the given sets of numbers are all the numbers solutions of 4x+7≠23 ?

(1 point)
Responses

3, 4, 6, 7
3, 4, 6, 7

2, 3, 5, 6
2, 3, 5, 6

4, 5, 6, 7
4, 5, 6, 7

1, 2, 3, 4

1 answer

To determine which set of numbers contains only solutions to the inequality \( 4x + 7 \neq 23 \), we first need to solve the equation \( 4x + 7 = 23 \) to find the value of \( x \) that does not satisfy the inequality.

  1. Solve the equation: \[ 4x + 7 = 23 \] Subtract 7 from both sides: \[ 4x = 16 \] Divide by 4: \[ x = 4 \]

This means the inequality \( 4x + 7 \neq 23 \) is satisfied for all values except \( x = 4 \).

Now, let's check each set of numbers to see which one does not include \( 4 \):

  1. Set 1: \( 3, 4, 6, 7 \)
    Contains \( 4 \) (not a solution).

  2. Set 2: \( 2, 3, 5, 6 \)
    Does not contain \( 4 \) (all are solutions).

  3. Set 3: \( 4, 5, 6, 7 \)
    Contains \( 4 \) (not a solution).

  4. Set 4: \( 1, 2, 3, 4 \)
    Contains \( 4 \) (not a solution).

Conclusion: The only set of numbers where all the numbers are solutions to the inequality \( 4x + 7 \neq 23 \) is Set 2: \( 2, 3, 5, 6 \).