To solve the system of inequalities, let's work through each inequality step-by-step.
### Inequality 1: [tex]\(6x - 21 \geq -3\)[/tex]
1. Add 21 to both sides to isolate the term containing [tex]\(x\)[/tex]:
[tex]\[
6x - 21 + 21 \geq -3 + 21
\][/tex]
Simplifying, we get:
[tex]\[
6x \geq 18
\][/tex]
2. Divide both sides by 6:
[tex]\[
\frac{6x}{6} \geq \frac{18}{6}
\][/tex]
Simplifying, we obtain:
[tex]\[
x \geq 3
\][/tex]
### Inequality 2: [tex]\(14x + 11 < -17 \)[/tex]
1. Subtract 11 from both sides to isolate the term containing [tex]\(x\)[/tex]:
[tex]\[
14x + 11 - 11 < -17 - 11
\][/tex]
Simplifying, we get:
[tex]\[
14x < -28
\][/tex]
2. Divide both sides by 14:
[tex]\[
\frac{14x}{14} < \frac{-28}{14}
\][/tex]
Simplifying, we obtain:
[tex]\[
x < -2
\][/tex]
### Intersection of the Solution Sets
Now, let's find the intersection of the solution sets from both inequalities:
- From the first inequality, we have [tex]\(x \geq 3\)[/tex].
- From the second inequality, we have [tex]\(x < -2\)[/tex].
There is no value of [tex]\(x\)[/tex] that can simultaneously satisfy [tex]\(x \geq 3\)[/tex] and [tex]\(x < -2\)[/tex]. Therefore, the solution set is the empty set.
### Conclusion
There are no real numbers [tex]\(x\)[/tex] that simultaneously satisfy both inequalities. Thus, the solution is:
[tex]\[
\boxed{\text{There are no such real numbers.}}
\][/tex]
