In each of the inequalities below, solve for $$ x $$ and show the solution on a number line. 1. $$ -15 \leq 4 x+1

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In each of the inequalities below, solve for $$ x $$ and show the solution on a number line. 1. $$ -15 \leq 4 x+1<1 $$ and $$ x \in \mathbb{R} $$ 2. $$ -15<1-4 x<5 $$ and $$ x \in \mathbb{R} $$ 3. $$ 5<2 x+3 \leq 17 $$ and $$ x \in \mathbb{R} $$ 4. $$ -5<2 x-3 \leq 11 $$ and $$ x \in \mathbb{Z} $$ 5. $$ -12 \leq 3-3 x<6 $$ and $$ x \in \mathbb{N} $$ 6. $$ \frac{1}{2}<\frac{x-1}{6} \leq \frac{1}{12} $$ and $$ x \in \mathbb{Z} $$ 7. $$ -2 \leq 2-4 x<6 $$ and $$ x \in \mathbb{R} $$ 8. $$ 0<\frac{x}{3}+1 \leq 3 $$ and $$ x \in \mathbb{R} $$ 9. $$ \frac{1}{4} \leq \frac{x}{12}+\frac{1}{3}<1 \frac{1}{6} $$ and $$ x \in \mathbb{Z} $$

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1. Solve $$ -15 \le 4x+1 < 1. $$ • Subtract $$1$$ from all parts: $$ -15-1 \le 4x < 1-1 \quad \Longrightarrow \quad -16 \le 4x < 0. $$ • Divide every part by $$4$$ (note that $$4>0$$ so the inequalities do not flip): $$ \frac{-16}{4} \le x < \frac{0}{4} \quad \Longrightarrow \quad -4 \le x < 0. $$ **Number Line:** A closed circle at $$-4$$ and an open circle at $$0$$, with shading in between. 2. Solve $$ -15 < 1-4x < 5. $$ • Subtract $$1$$ from all parts: $$ -15-1 < -4x < 5-1 \quad \Longrightarrow \quad -16 < -4x < 4. $$ • Divide every part by $$-4$$ (this reverses the inequality signs since $$-4<0$$): $$ \frac{-16}{-4} > x > \frac{4}{-4} \quad \Longrightarrow \quad 4 > x > -1. $$ This can be equivalently written as $$ -1 < x < 4. $$ **Number Line:** An open circle at $$-1$$ and an open circle at $$4$$, shading between them. 3. Solve $$ 5 < 2x+3 \le 17. $$ • Subtract $$3$$ from all parts: $$ 5-3 < 2x \le 17-3 \quad \Longrightarrow \quad 2 < 2x \le 14. $$ • Divide by $$2$$: $$ \frac{2}{2} < x \le \frac{14}{2} \quad \Longrightarrow \quad 1 < x \le 7. $$ **Number Line:** An open circle at $$1$$ and a closed circle at $$7$$, shading the interval in between. 4. Solve $$ -5 < 2x-3 \le 11, \quad x \in \mathbb{Z}. $$ • Add $$3$$ to each part: $$ -5+3 < 2x \le 11+3 \quad \Longrightarrow \quad -2 < 2x \le 14. $$ • Divide by $$2$$: $$ \frac{-2}{2} < x \le \frac{14}{2} \quad \Longrightarrow \quad -1 < x \le 7. $$ Since $$x$$ must be an integer and $$x > -1$$, the smallest integer is $$0$$. Thus, $$ x \in \{0, 1, 2, 3, 4, 5, 6, 7\}. $$ **Number Line:** Place integers $$0$$ through $$7$$ on the number line with each point marked. 5. Solve $$ -12 \le 3-3x < 6, \quad x \in \mathbb{N}. $$ • Subtract $$3$$ from all parts: $$ -12-3 \le -3x < 6-3 \quad \Longrightarrow \quad -15 \le -3x < 3. $$ • Divide by $$-3$$ (remember to reverse the inequalities): $$ \frac{-15}{-3} \ge x > \frac{3}{-3} \quad \Longrightarrow \quad 5 \ge x > -1. $$ Interpreting $$x \in \mathbb{N}$$ as the set of positive integers, we have: $$ x \in \{1, 2, 3, 4, 5\}. $$ **Number Line:** Mark the natural numbers $$1$$ through $$5$$. 6. Solve $$ \frac{1}{2} < \frac{x-1}{6} \le \frac{1}{12}, \quad x \in \mathbb{Z}. $$ • Multiply every part by $$6$$: $$ 6\cdot\frac{1}{2} < x-1 \le 6\cdot\frac{1}{12} \quad \Longrightarrow \quad 3 < x-1 \le \frac{1}{2}. $$ • Add $$1$$ to all parts: $$ 3+1 < x \le \frac{1}{2}+1 \quad \Longrightarrow \quad 4 < x \le \frac{3}{2}. $$ The inequality $$4 < x \le \frac{3}{2}$$ is impossible since $$4 > \frac{3}{2}$$. **Thus, there is no solution.** 7. Solve $$ -2 \le 2-4x < 6. $$ • Subtract $$2$$ from each part: $$ -2-2 \le 2-4x-2 < 6-2 \quad \Longrightarrow \quad -4 \le -4x < 4. $$ • Divide by $$-4$$ (remembering to reverse the inequalities): $$ \frac{-4}{-4} \ge x > \frac{4}{-4} \quad \Longrightarrow \quad 1 \ge x > -1. $$ This can be written as: $$ -1 < x \le 1. $$ **Number Line:** An open circle at $$-1$$ and a closed circle at $$1$$, with shading between. 8. Solve $$ 0 < \frac{x}{3}+1 \le 3. $$ • Subtract $$1$$ from each part: $$ 0-1 < \frac{x}{3} \le 3-1 \quad \Longrightarrow \quad -1 < \frac{x}{3} \le 2. $$ • Multiply by $$3$$: $$ -3 < x \le 6. $$ **Number Line:** An open circle at $$-3$$ and a closed circle at $$6$$, shading the interval in between. 9. Solve $$ \frac{1}{4} \le \frac{x}{12}+\frac{1}{3} < 1\frac{1}{6}, \quad x \in \mathbb{Z}. $$ First, rewrite $$1\frac{1}{6}$$ as an improper fraction: $$ 1\frac{1}{6} = \frac{7}{6}. $$ So the inequality becomes: $$ \frac{1}{4} \le \frac{x}{12}+\frac{1}{3} < \frac{7}{6}. $$ • Multiply every part by $$12$$ to eliminate denominators: $$ 12\cdot\frac{1}{4} \le x + 12\cdot\frac{1}{3} < 12\cdot\frac{7}{6}. $$ Evaluate each term: $$ 3 \le x + 4 < 14. $$ • Subtract $$4$$ from each part: $$ 3-4 \le x < 14-4 \quad \Longrightarrow \quad -1 \le x < 10. $$ Since $$x \in \mathbb{Z}$$, the solutions are: $$ x \in \{-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}. $$ **Number Line:** Mark all integers from $$-1$$ up to $$9$$ (closed at $$-1$$ and an open point at $$10$$). 1. $$ -4 \le x < 0 $$ 2. $$ -1 < x < 4 $$ 3. $$ 1 < x \le 7 $$ 4. $$ x = 0, 1, 2, 3, 4, 5, 6, 7 $$ 5. $$ x = 1, 2, 3, 4, 5 $$ 6. No solution 7. $$ -1 < x \le 1 $$ 8. $$ -3 < x \le 6 $$ 9. $$ x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 $$

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