Question

1. $|3x + 3| leq -5$ 19. $|6y - 1|>3$ 22. $|3 - 2t| geq -3$

Explanation:

Step1: Solve \(|3x + 3|\leq - 5\)

The absolute - value of any real number \(a\), denoted as \(|a|\), is always non - negative, i.e., \(|a|\geq0\) for all real numbers \(a\). Since \(|3x + 3|\geq0\) for all real \(x\), the inequality \(|3x + 3|\leq - 5\) has no solution because a non - negative number cannot be less than or equal to a negative number.

Step2: Solve \(|6y-1|>3\)

We consider two cases:
Case 1: \(6y - 1>3\). Add 1 to both sides: \(6y>3 + 1\), so \(6y>4\). Then divide both sides by 6: \(y>\frac{4}{6}=\frac{2}{3}\).
Case 2: \((6y - 1)<-3\). Add 1 to both sides: \(6y<-3 + 1\), so \(6y<-2\). Then divide both sides by 6: \(y<-\frac{2}{6}=-\frac{1}{3}\).
So the solution of \(|6y - 1|>3\) is \(y<-\frac{1}{3}\) or \(y>\frac{2}{3}\).

Step3: Solve \(|3-2t|\geq - 3\)

Since the absolute - value of any real number is non - negative, \(|3-2t|\geq0\) for all real \(t\). And since \(0\geq - 3\), the inequality \(|3-2t|\geq - 3\) is true for all real numbers \(t\), i.e., the solution set is \((-\infty,\infty)\).

Answer:

1. No solution.
2. \(y<-\frac{1}{3}\) or \(y>\frac{2}{3}\).
3. All real numbers \(t\) (or \((-\infty,\infty)\)).