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Solve the inequality.
[tex]\[
\begin{array}{c}
6x - 2 \leq 9 \text{ or } 4 + 3x \ \textgreater \ 15 \\
x \leq \frac{11}{6} \text{ or } x \ \textgreater \ \frac{11}{3}
\end{array}
\][/tex]
### First Inequality: [tex]\( 6x - 2 \leq 9 \)[/tex]
1. Add 2 to both sides to isolate the term involving [tex]\( x \)[/tex]: [tex]\[
6x - 2 + 2 \leq 9 + 2
\][/tex] [tex]\[
6x \leq 11
\][/tex] 2. Divide both sides by 6 to solve for [tex]\( x \)[/tex]: [tex]\[
x \leq \frac{11}{6}
\][/tex]
### Second Inequality: [tex]\( 4 + 3x > 15 \)[/tex]
1. Subtract 4 from both sides to isolate the term involving [tex]\( x \)[/tex]: [tex]\[
4 + 3x - 4 > 15 - 4
\][/tex] [tex]\[
3x > 11
\][/tex] 2. Divide both sides by 3 to solve for [tex]\( x \)[/tex]: [tex]\[
x > \frac{11}{3}
\][/tex]
### Combined Solution
The solution to the compound inequality [tex]\( 6x - 2 \leq 9 \)[/tex] or [tex]\( 4 + 3x > 15 \)[/tex] is:
[tex]\[
x \leq \frac{11}{6} \quad \text{or} \quad x > \frac{11}{3}
\][/tex]
So, in the format provided in the question:
[tex]\[
x \leq \frac{11}{6} \text{ or } x > \frac{11}{3}
\][/tex]
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