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

? (1 point)
Responses

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

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

1, 2, 3, 4
1, 2, 3, 4

3, 4, 6, 7

1 answer

To determine which set of numbers contains all solutions to the inequality \(4x + 7 \neq 23\), we first need to solve the equation \(4x + 7 = 23\) to identify the values of \(x\) that do not satisfy the inequality.

  1. Start from the equation: \[ 4x + 7 = 23 \]

  2. Subtract 7 from both sides: \[ 4x = 23 - 7 \] \[ 4x = 16 \]

  3. Divide both sides by 4: \[ x = 4 \]

The only solution to the equation \(4x + 7 = 23\) is \(x = 4\). Therefore, the inequality \(4x + 7 \neq 23\) holds for all numbers except \(x = 4\).

Now we can examine each of the provided sets of numbers to see which set does not include the number 4 (as all the numbers in that set would satisfy the inequality):

  • Set 1: \(4, 5, 6, 7\) → Includes 4 (not a solution).
  • Set 2: \(2, 3, 5, 6\) → Does not include 4 (all are valid solutions).
  • Set 3: \(1, 2, 3, 4\) → Includes 4 (not a solution).
  • Set 4: \(3, 4, 6, 7\) → Includes 4 (not a solution).

The only set that contains numbers all of which are solutions to \(4x + 7 \neq 23\) is:

2, 3, 5, 6