Answer :
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]
### 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]
