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

Question

(1 point)
Responses

8, 7, 6
8, 7, 6

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

0, 3, 9, 12
0, 3, 9, 12

9, 10, 11, 12

Answer 1

To determine which set of numbers contains only solutions to the inequality \(18 + 2x \geq 36\), we first need to solve the inequality.

Starting with the inequality:

\[ 18 + 2x \geq 36 \]

Subtract 18 from both sides:

\[ 2x \geq 36 - 18 \]

\[ 2x \geq 18 \]

Now, divide both sides by 2:

\[ x \geq 9 \]

This means that any number that is a solution to the inequality must be 9 or greater.

Now, let's examine each of the given sets of numbers:

1. Set: 8, 7, 6

   • All numbers are less than 9. (Not valid)

2. Set: 9, 8, 7, 6

   • Contains numbers 8, 7, 6 which are less than 9. (Not valid)

3. Set: 0, 3, 9, 12

   • Contains numbers 0, 3 which are less than 9. (Not valid)

4. Set: 9, 10, 11, 12

   • All numbers are 9 or greater. (Valid)

The only set where all numbers are solutions to the inequality \(18 + 2x \geq 36\) is:

9, 10, 11, 12