Find all real numbers [tex]x[/tex] such that [tex]4x + 2 \ \textgreater \ 14[/tex] and [tex]-21x + 1 \ \textgreater \ 22[/tex].

Click on the correct answer.

A. [tex]x \ \textgreater \ 3.5[/tex] and [tex]x \ \textless \ -1[/tex]

B. [tex]x \ \textgreater \ 3.5[/tex] and [tex]x \ \textless \ -1/21[/tex]

C. [tex]x \ \textgreater \ 3[/tex] and [tex]x \ \textless \ -1[/tex]

D. [tex]x \ \textgreater \ 3[/tex] and [tex]x \ \textless \ -1/21[/tex]



Answer :

To solve this problem, we need to address each inequality individually and then combine their solutions. Here are the steps:

### Step 1: Solve the first inequality [tex]\(4x + 2 > 14\)[/tex]

1. Subtract 2 from both sides of the inequality:
[tex]\[ 4x + 2 - 2 > 14 - 2 \][/tex]
Simplifies to:
[tex]\[ 4x > 12 \][/tex]
2. Divide both sides by 4:
[tex]\[ \frac{4x}{4} > \frac{12}{4} \][/tex]
Simplifies to:
[tex]\[ x > 3 \][/tex]

### Step 2: Solve the second inequality [tex]\(-21x + 1 > 22\)[/tex]

1. Subtract 1 from both sides of the inequality:
[tex]\[ -21x + 1 - 1 > 22 - 1 \][/tex]
Simplifies to:
[tex]\[ -21x > 21 \][/tex]
2. Divide both sides by -21. Note that dividing by a negative number reverses the inequality sign:
[tex]\[ \frac{-21x}{-21} < \frac{21}{-21} \][/tex]
Simplifies to:
[tex]\[ x < -1 \][/tex]

### Combining the solutions:
- From the first inequality, we have [tex]\(x > 3\)[/tex].
- From the second inequality, we have [tex]\(x < -1\)[/tex].

Thus, we need [tex]\(x\)[/tex] to be both greater than 3 and less than -1 simultaneously.

### Conclusion:
There are no real numbers [tex]\(x\)[/tex] that satisfy both inequalities at the same time. Therefore, the solution set is the empty set.

[tex]\[ \boxed{\text{No real numbers } x \text{ satisfy both inequalities}} \][/tex]