In which of the given sets of numbers are all the numbers solutions of 18+2x≥36

Question

?(1 point)
Responses

9, 10, 11, 12
9, 10, 11, 12

8, 7, 6
8, 7, 6

9, 8, 7, 6
9, 8, 7, 6

0, 3, 9, 12

Answer 1

To solve the inequality \(18 + 2x \geq 36\), we first isolate \(x\):

1. Subtract 18 from both sides: \[ 2x \geq 36 - 18 \] \[ 2x \geq 18 \]

2. Now divide both sides by 2: \[ x \geq \frac{18}{2} \] \[ x \geq 9 \]

This means that any value of \(x\) must be greater than or equal to 9.

Now, let's evaluate each set of numbers:

1. Set 1: 9, 10, 11, 12

   • All values (9, 10, 11, 12) are \(\geq 9\).
   • This set is a solution.

2. Set 2: 8, 7, 6

   • None of these values (8, 7, 6) are \(\geq 9\).
   • This set is not a solution.

3. Set 3: 9, 8, 7, 6

   • Only 9 is \(\geq 9\) (8, 7, 6 are not).
   • This set is not a solution.

4. Set 4: 0, 3, 9, 12

   • Only 9 and 12 are \(\geq 9\) (0 and 3 are not).
   • This set is not a solution.

Based on this analysis, the only set where all numbers are solutions of the inequality \(18 + 2x \geq 36\) is:

Response 1: 9, 10, 11, 12.