Solve for k.

Question

k+14≤12 or 3k–2≥10
Write your answer as a compound inequality with integers.

Answer 1

To solve the compound inequality $k+14\le 12$ or $3k-2\ge 10$, we will solve each inequality separately.

First Inequality: $$k+14\le 12$$

Subtract 14 from both sides: $$k\le 12-14$$ $$k\le -2$$

Second Inequality: $$3k-2\ge 10$$

Add 2 to both sides: $$3k\ge 10+2$$ $$3k\ge 12$$

Now divide by 3: $$k\ge \frac{12}{3}$$ $$k\ge 4$$

Combining the Solutions: The solutions from the two inequalities are:

• $k\le -2$
• $k\ge 4$

Thus, the final solution can be written as a compound inequality: $$k\le -2\text{or}k\ge 4$$

In interval notation, this can also be represented as: $$(-\mathrm{\infty },-2]\cup [4,\mathrm{\infty })$$

So the answer is: $$k\le -2\text{or}k\ge 4$$