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]
