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Solve the compound inequality for [tex]\( x \)[/tex]:

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

Select one:
A. [tex]\( x \ \textgreater \ \frac{16}{6} \)[/tex] or [tex]\( x \ \textless \ \frac{1}{3} \)[/tex]
B. [tex]\( x \ \textgreater \ 2 \)[/tex] or [tex]\( x \ \textless \ 1 \)[/tex]
C. [tex]\(-2 \ \textless \ x \ \textgreater \ 1 \)[/tex]
D. [tex]\( x \ \textgreater \ -2 \)[/tex] and [tex]\( x \ \textless \ -1 \)[/tex]



Answer :

GinnyAnswer
To solve the compound inequality [tex]\( 6x + 2 > 14 \text{ or } 3x - 1 < 2 \)[/tex], you need to solve each inequality separately and then combine their results logically.

1. Solve the inequality [tex]\( 6x + 2 > 14 \)[/tex]:
[tex]\[ 6x + 2 > 14 \][/tex]
Subtract 2 from both sides:
[tex]\[ 6x > 12 \][/tex]
Divide both sides by 6:
[tex]\[ x > 2 \][/tex]

2. Solve the inequality [tex]\( 3x - 1 < 2 \)[/tex]:
[tex]\[ 3x - 1 < 2 \][/tex]
Add 1 to both sides:
[tex]\[ 3x < 3 \][/tex]
Divide both sides by 3:
[tex]\[ x < 1 \][/tex]

Now, combine the results of the two inequalities:
- From the first inequality, we have [tex]\( x > 2 \)[/tex].
- From the second inequality, we have [tex]\( x < 1 \)[/tex].

Therefore, the solution to the compound inequality [tex]\( 6x + 2 > 14 \text{ or } 3x - 1 < 2 \)[/tex] is:
- [tex]\( x > 2 \)[/tex] or [tex]\( x < 1 \)[/tex].

So, the correct answer is:
b. [tex]\( x > 2 \)[/tex] or [tex]\( x < 1 \)[/tex].

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