Find all real numbers [tex]\( x \)[/tex] such that [tex]\( 6x - 21 \leq -3 \)[/tex] or [tex]\( 14x + 11 \leq -17 \)[/tex].

Click on the correct answer.



Answer :

To find all real numbers [tex]\( x \)[/tex] such that [tex]\( 6x - 21 \leq -3 \)[/tex] or [tex]\( 14x + 11 \leq -17 \)[/tex], we will solve each inequality separately and then combine the results.

1. Solve the first inequality: [tex]\( 6x - 21 \leq -3 \)[/tex]

- Add 21 to both sides:
[tex]\[ 6x - 21 + 21 \leq -3 + 21 \][/tex]
[tex]\[ 6x \leq 18 \][/tex]

- Divide both sides by 6:
[tex]\[ x \leq 3 \][/tex]

2. Solve the second inequality: [tex]\( 14x + 11 \leq -17 \)[/tex]

- Subtract 11 from both sides:
[tex]\[ 14x + 11 - 11 \leq -17 - 11 \][/tex]
[tex]\[ 14x \leq -28 \][/tex]

- Divide both sides by 14:
[tex]\[ x \leq -2 \][/tex]

3. Combine the solutions:

For [tex]\( x \)[/tex] to satisfy the original statement [tex]\( 6x - 21 \leq -3 \)[/tex] or [tex]\( 14x + 11 \leq -17 \)[/tex], it must satisfy either of the individual inequalities.

- From the first inequality, we have [tex]\( x \leq 3 \)[/tex].
- From the second inequality, we have [tex]\( x \leq -2 \)[/tex].

Since any number that satisfies [tex]\( x \leq -2 \)[/tex] also satisfies [tex]\( x \leq 3 \)[/tex], the more restrictive condition [tex]\( x \leq -2 \)[/tex] is sufficient to describe the complete solution.

Therefore, the combined solution for the inequalities is:

[tex]\[ x \leq 3 \][/tex]

In conclusion, the correct answer is all real numbers [tex]\( x \)[/tex] such that:

[tex]\[ x \leq 3 \][/tex]