To solve the given problem, we need to solve the inequalities separately and then determine the union of their solution sets. Let's start with each inequality step-by-step.
### Solving the first inequality: [tex]\( 6x - 4 \leq -40 \)[/tex]
1. Add 4 to both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[
6x - 4 + 4 \leq -40 + 4
\][/tex]
[tex]\[
6x \leq -36
\][/tex]
2. Divide both sides by 6 to solve for [tex]\( x \)[/tex]:
[tex]\[
\frac{6x}{6} \leq \frac{-36}{6}
\][/tex]
[tex]\[
x \leq -6
\][/tex]
So, the solution to the inequality [tex]\( 6x - 4 \leq -40 \)[/tex] is:
[tex]\[
x \leq -6
\][/tex]
### Solving the second inequality: [tex]\( \frac{x}{16} \geq 1 \)[/tex]
1. Multiply both sides by 16 to isolate [tex]\( x \)[/tex]:
[tex]\[
\frac{x}{16} \times 16 \geq 1 \times 16
\][/tex]
[tex]\[
x \geq 16
\][/tex]
So, the solution to the inequality [tex]\( \frac{x}{16} \geq 1 \)[/tex] is:
[tex]\[
x \geq 16
\][/tex]
### Combining the solutions
Since the original problem uses the logical "or", we combine the solutions of the two inequalities.
The solution set for the inequality [tex]\( 6x - 4 \leq -40 \)[/tex] is:
[tex]\[
x \leq -6
\][/tex]
And the solution set for the inequality [tex]\( \frac{x}{16} \geq 1 \)[/tex] is:
[tex]\[
x \geq 16
\][/tex]
The union of these two sets is:
[tex]\[
x \leq -6 \quad \text{or} \quad x \geq 16
\][/tex]
This represents all [tex]\( x \)[/tex] that are either less than or equal to -6, or greater than or equal to 16. In interval notation, this can be written as:
[tex]\[
(-\infty, -6] \cup [16, \infty)
\][/tex]
Thus, the solution to the problem is:
[tex]\[ x \leq -6 \quad \text{or} \quad x \geq 16. \][/tex]
