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

Question

? (1 point)
Responses

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

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

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

1, 2, 3, 4

Answer 1

To determine which of the given sets of numbers are all solutions of the inequality \(4x + 7 \neq 23\), we first need to solve the equation \(4x + 7 = 23\).

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

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

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

This indicates that \(x = 4\) is the only value that makes the equation true, so the inequality \(4x + 7 \neq 23\) means that \(x\) cannot be equal to 4.

Now, let's evaluate each set of numbers to see if they contain the number 4.

• Set 1: \( 2, 3, 5, 6 \) - None of these are 4. All numbers are solutions.
• Set 2: \( 4, 5, 6, 7 \) - Contains 4. Not all numbers are solutions.
• Set 3: \( 3, 4, 6, 7 \) - Contains 4. Not all numbers are solutions.
• Set 4: \( 1, 2, 3, 4 \) - Contains 4. Not all numbers are solutions.

Thus, the only set in which all the numbers are solutions of \(4x + 7 \neq 23\) is:

2, 3, 5, 6.