Question- Q20: Solve and express in interval notation: $|3 x+1| \geq 4$. - Q21: Solve and express in interval notation: $|2 x-7|>3$.

STEP 1

1. We are dealing with absolute value inequalities.
2. For an inequality of the form $|A| \geq B$, the solution is $A \geq B$ or $A \leq -B$.
3. For an inequality of the form $|A| > B$, the solution is $A > B$ or $A < -B$.
### _HIGH_LEVEL_APPROACH_ for Q20:
1. Set up the inequality for $|3x + 1| \geq 4$.
2. Solve the inequality $3x + 1 \geq 4$.
3. Solve the inequality $3x + 1 \leq -4$.
4. Combine the solutions and express in interval notation.
### _HIGH_LEVEL_APPROACH_ for Q21:
1. Set up the inequality for $|2x - 7| > 3$.
2. Solve the inequality $2x - 7 > 3$.
3. Solve the inequality $2x - 7 < -3$.
4. Combine the solutions and express in interval notation.
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## Q20: Solve and express in interval notation: $|3x + 1| \geq 4$.

STEP 2

STEP 3

Set up the two inequalities:
$$3x + 1 \geq 4 \quad \text{or} \quad 3x + 1 \leq -4$$

STEP 4

Solve the inequality $3x + 1 \geq 4$:
$$3x + 1 \geq 4$$ $$3x \geq 3$$ $$x \geq 1$$

STEP 5

Solve the inequality $3x + 1 \leq -4$:
$$3x + 1 \leq -4$$ $$3x \leq -5$$ $$x \leq -\frac{5}{3}$$

STEP 6

Combine the solutions from STEP_2 and STEP_3. The solution in interval notation is:
$$x \in (-\infty, -\frac{5}{3}] \cup [1, \infty)$$
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## Q21: Solve and express in interval notation: $|2x - 7| > 3$.
STEP_1: Set up the two inequalities:
$$2x - 7 > 3 \quad \text{or} \quad 2x - 7 < -3$$
STEP_2: Solve the inequality $2x - 7 > 3$:
$$2x - 7 > 3$$ $$2x > 10$$ $$x > 5$$
STEP_3: Solve the inequality $2x - 7 < -3$:
$$2x - 7 < -3$$ $$2x < 4$$ $$x < 2$$
STEP_4: Combine the solutions from STEP_2 and STEP_3. The solution in interval notation is:
$$x \in (-\infty, 2) \cup (5, \infty)$$
The solutions are: - For Q20: $x \in (-\infty, -\frac{5}{3}] \cup [1, \infty)$ - For Q21: $x \in (-\infty, 2) \cup (5, \infty)$