Question

I which of the given sets of numbers are all the numbers solutions of 18 + 2x >36?
A. 0, 3, 9, 12
B. 9, 8, 7, 6
C. 9, 10, 11, 12
D. 8, 7, 6

Answer 1

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

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

2. Divide both sides by 2:
\[
x > 9
\]

Now we look for the sets of numbers where all members are greater than 9.

A. \( 0, 3, 9, 12 \) → Not all are greater than 9 (0, 3, and 9 are not greater than 9).
B. \( 9, 8, 7, 6 \) → Not all are greater than 9 (9, 8, 7, and 6 are not greater than 9).
C. \( 9, 10, 11, 12 \) → Only 10, 11, and 12 are greater than 9 (9 is not greater than 9).
D. \( 8, 7, 6 \) → Not all are greater than 9 (all are not greater than 9).

None of the sets have numbers that are all greater than 9.

However, set C, which includes numbers larger than 9, only fails because it contains 9 itself.

**The answer is that no option fits the requirement that all numbers must be greater than 9.**