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]



Answer :

Let's solve each inequality step-by-step.

### 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]