Relational Signs & Inequalities

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1. Basic Relational Signs Less Than: $a Example: $3 Greater Than: $a > b$ (a is strictly larger than b) Example: $7 > 2$, $-1 > -5$, $10 > 0$ Less Than or Equal To: $a \le b$ (a is smaller than or equal to b) Example: $3 \le 5$, $3 \le 3$, $-4 \le 0$ Greater Than or Equal To: $a \ge b$ (a is larger than or equal to b) Example: $8 \ge 1$, $8 \ge 8$, $0 \ge -2$ Equal To: $a = b$ (a is exactly equal to b) Example: $5 = 5$, $2+3 = 5$ Not Equal To: $a \ne b$ (a is not equal to b) Example: $5 \ne 7$, $1 \ne 0$ 2. Fundamental Properties of Inequalities 2.1. Addition/Subtraction Property If $a Example: Given $3 If $a Example: Given $10 > 4$. Subtract $3$ from both sides: $10-3 > 4-3 \implies 7 > 1$. (True) Rule: Adding or subtracting the same number from both sides of an inequality does not change the direction of the inequality sign. 2.2. Multiplication/Division Property (Positive Number) If $a 0$, then $ac Example: Given $2 0$): $2 \times 3 If $a 0$, then $\frac{a}{c} Example: Given $10 > 4$. Divide by $2$ ($c=2 > 0$): $\frac{10}{2} > \frac{4}{2} \implies 5 > 2$. (True) Rule: Multiplying or dividing both sides of an inequality by the same positive number does not change the direction of the inequality sign. 2.3. Multiplication/Division Property (Negative Number) If $a bc$ Example: Given $2 5 \times (-1) \implies -2 > -5$. (Sign reversed, True) If $a \frac{b}{c}$ Example: Given $6 > 3$. Divide by $-3$ ($c=-3 Rule: Multiplying or dividing both sides of an inequality by the same negative number reverses the direction of the inequality sign. 2.4. Transitive Property If $a Example: If $2 If $a > b$ and $b > c$, then $a > c$ Example: If $10 > 6$ and $6 > 1$, then $10 > 1$. (True) 2.5. Reciprocal Property If $a > 0$ and $b > 0$: If $a \frac{1}{b}$ Example: Given $2 \frac{1}{5}$ (True, $0.5 > 0.2$) If $a \frac{1}{b}$ Example: Given $-5 -\frac{1}{2}$ (True, $-0.2 > -0.5$) Rule: Taking the reciprocal of both sides of an inequality (when both sides have the same sign) reverses the inequality sign. Caution: This property does not hold if $a$ and $b$ have different signs (e.g., $-2 3. Solving Linear Inequalities Treat inequalities much like equations, applying the properties above. Key Difference: Remember to reverse the sign if multiplying or dividing by a negative number. Example Walkthrough: Solve $3x - 4 Original inequality: $3x - 4 Subtract $5x$ from both sides: $3x - 5x - 4 Add $4$ to both sides: $-2x - 4 + 4 Divide by $-2$ (a negative number). Reverse the inequality sign: $\frac{-2x}{-2} > \frac{6}{-2} \implies x > -3$ Solution: $x > -3$ 4. Solving Compound Inequalities "And" Inequalities: $a a$ AND $x Solve each part separately and find the intersection of the solutions. Example: Solve $-2 \le 2x + 4 Solve $-2 \le 2x + 4$: $-2 - 4 \le 2x \implies -6 \le 2x \implies -3 \le x$ Solve $2x + 4 Combine solutions: We need both $x \ge -3$ AND $x Interval Notation: $[-3, 3)$ "Or" Inequalities: $x b$ Solve each part separately and combine the solutions (union). Example: Solve $x + 1 5$ Solve $x + 1 Solve $x - 3 > 5$: $x > 5 + 3 \implies x > 8$ Combine solutions: $x 8$. These are two separate regions on the number line. Interval Notation: $(-\infty, -3) \cup (8, \infty)$ 5. Absolute Value Inequalities Type 1: $|X| 0$) This means $-a Example: Solve $|2x - 1| $-5 Add $1$ to all parts: $-5 + 1 Divide by $2$: $\frac{-4}{2} Solution: $(-2, 3)$ Type 2: $|X| > a$ (where $a > 0$) This means $X a$. Example: Solve $|3x + 2| \ge 7$ $3x + 2 \le -7$ OR $3x + 2 \ge 7$ Solve $3x + 2 \le -7$: $3x \le -7 - 2 \implies 3x \le -9 \implies x \le -3$ Solve $3x + 2 \ge 7$: $3x \ge 7 - 2 \implies 3x \ge 5 \implies x \ge \frac{5}{3}$ Solution: $x \le -3$ OR $x \ge \frac{5}{3}$ Interval Notation: $(-\infty, -3] \cup [\frac{5}{3}, \infty)$ Special Cases: $|X| $|X| > -5$: All real numbers (absolute value is always greater than a negative number). $|X| $|X| \le 0$: $X=0$ (only 0 has absolute value 0). 6. Interval Notation (Summary) A concise way to express ranges of numbers, often used for inequality solutions. Inequality Interval Notation Description $a $(a, b)$ Open interval; $x$ is between $a$ and $b$, excluding $a, b$. $a \le x \le b$ $[a, b]$ Closed interval; $x$ is between $a$ and $b$, including $a, b$. $a $(a, b]$ Half-open; $x$ is greater than $a$ and less than or equal to $b$. $a \le x $[a, b)$ Half-open; $x$ is greater than or equal to $a$ and less than $b$. $x $(-\infty, a)$ All numbers less than $a$. $x \le a$ $(-\infty, a]$ All numbers less than or equal to $a$. $x > a$ $(a, \infty)$ All numbers greater than $a$. $x \ge a$ $[a, \infty)$ All numbers greater than or equal to $a$. All Real Numbers $(-\infty, \infty)$ The entire number line. 7. Quadratic Inequalities Solving inequalities involving quadratic expressions like $ax^2 + bx + c > 0$. Find Critical Points: Set the quadratic expression equal to zero and solve for $x$. These are the roots of the quadratic equation. Plot Critical Points: Place these points on a number line. They divide the line into intervals. Test Intervals: Pick a test value from each interval and substitute it into the original inequality. Determine Solution: Identify the intervals where the inequality holds true. Example: Solve $x^2 - x - 6 > 0$ Critical Points: $x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0$. Roots are $x=3$ and $x=-2$. Intervals: $(-\infty, -2)$, $(-2, 3)$, $(3, \infty)$. Test Points: For $(-\infty, -2)$, pick $x=-3$: $(-3)^2 - (-3) - 6 = 9 + 3 - 6 = 6$. Since $6 > 0$, this interval is a solution. For $(-2, 3)$, pick $x=0$: $(0)^2 - (0) - 6 = -6$. Since $-6 \not> 0$, this interval is not a solution. For $(3, \infty)$, pick $x=4$: $(4)^2 - (4) - 6 = 16 - 4 - 6 = 6$. Since $6 > 0$, this interval is a solution. Solution: $x 3$. Interval Notation: $(-\infty, -2) \cup (3, \infty)$. 8. Rational Inequalities Solving inequalities with rational expressions, e.g., $\frac{P(x)}{Q(x)} > 0$. Find Critical Points: Set both the numerator and denominator equal to zero and solve for $x$. Plot Critical Points: Place these points on a number line. Remember that values making the denominator zero are excluded (represented by open circles). Test Intervals: Pick a test value from each interval and substitute it into the original inequality. Determine Solution: Identify the intervals where the inequality holds true. Example: Solve $\frac{x-1}{x+2} \le 0$ Critical Points: Numerator: $x-1 = 0 \implies x=1$ Denominator: $x+2 = 0 \implies x=-2$ (This value is excluded from the solution set) Intervals: $(-\infty, -2)$, $(-2, 1]$, $[1, \infty)$. Note $x=1$ is included because of "$\le$". Test Points: For $(-\infty, -2)$, pick $x=-3$: $\frac{-3-1}{-3+2} = \frac{-4}{-1} = 4$. Since $4 \not\le 0$, not a solution. For $(-2, 1)$, pick $x=0$: $\frac{0-1}{0+2} = \frac{-1}{2} = -0.5$. Since $-0.5 \le 0$, this interval is a solution. For $(1, \infty)$, pick $x=2$: $\frac{2-1}{2+2} = \frac{1}{4} = 0.25$. Since $0.25 \not\le 0$, not a solution. Solution: $-2 Interval Notation: $(-2, 1]$.