Solve the compound inequality for [tex]$x$[/tex]:

[tex]\[6x + 2 \ \textgreater \ 14 \text{ or } 3x - 1 \ \textless \ 2\][/tex]

Select one:
a. [tex]$x \ \textgreater \ 2$[/tex] or [tex]$x \ \textless \ 1$[/tex]
b. [tex]-2 \ \textless \ x \ \textgreater \ 1$[/tex]
c. [tex]x \ \textgreater \ -2[/tex] and [tex]x \ \textless \ -1$[/tex]
d. [tex]x \ \textgreater \ \frac{16}{6}$[/tex] or [tex]x \ \textless \ \frac{1}{3}$[/tex]



Answer :

Certainly! Let's solve the given compound inequality step by step.

### Inequality 1: [tex]\(6x + 2 > 14\)[/tex]

1. Subtract 2 from both sides:
[tex]\[ 6x + 2 - 2 > 14 - 2 \][/tex]
Simplifying this, we get:
[tex]\[ 6x > 12 \][/tex]

2. Divide both sides by 6:
[tex]\[ \frac{6x}{6} > \frac{12}{6} \][/tex]
Simplifying further, we get:
[tex]\[ x > 2 \][/tex]

### Inequality 2: [tex]\(3x - 1 < 2\)[/tex]

1. Add 1 to both sides:
[tex]\[ 3x - 1 + 1 < 2 + 1 \][/tex]
Simplifying this, we get:
[tex]\[ 3x < 3 \][/tex]

2. Divide both sides by 3:
[tex]\[ \frac{3x}{3} < \frac{3}{3} \][/tex]
Simplifying further, we get:
[tex]\[ x < 1 \][/tex]

### Combining the Results:

The solution to the compound inequality [tex]\(6x + 2 > 14\)[/tex] or [tex]\(3x - 1 < 2\)[/tex] is:

[tex]\[ x > 2 \quad \text{or} \quad x < 1 \][/tex]

Hence, the correct choice is:
a. [tex]\(x > 2\)[/tex] or [tex]\(x < 1\)[/tex]