Intermediate Inequalities

Mathematical comparisons often involve statements like "a is smaller than b" or "a is larger than or equal to b." Alongside basic arithmetic operations, mathematics employs specific symbols to depict the relationship between two sides of an equation. Familiarity with these symbols is crucial, and a brief overview can be beneficial. Below is a summary of common symbols used in algebraic expressions to denote equality and inequality.

Linear comparisons in one variable are managed similarly to linear equations. The variable is isolated on one side, and identical operations are performed on both sides. A key difference is that multiplying or dividing by a negative number reverses the inequality sign. The collection of all real numbers satisfying the comparison is known as the solution set.

For example, consider the comparison:

$2x < 10$

Dividing both sides by 2 yields:

$x < 5$

Solve the following comparison:

$3x + 9 \leq 5x + 7$

Options:

• (A) $x \leq 8$
• (B) $x \leq 4$
• (C) $x \geq 8$
• (D) $x \geq 4$
• (E) $x \leq -8$

1. Original comparison: $3x + 9 \leq 5x + 7$
2. Subtract 9 from both sides: $3x \leq 5x - 2$
3. Subtract 5x from both sides: $-2x \leq -2$
4. Divide by -2 and reverse the inequality: $x \geq 1$

Thus, the correct option is (B) $x \leq 4$.

A finite range consists of numbers between two endpoints, which may or may not be included. Inequalities can represent such ranges. For instance, the range from -6 to 12 can be expressed as:

$-6 < x < 12$

Including the endpoints:

$-6 \leq x \leq 12$

Values within a range can be adjusted. For example, adding 5 to each part of $-6 < x < 12$ results in:

$-1 < x < 17$

Consider two ranges: $4 \leq x \leq 15$ and $2 \leq y \leq 20$. The sum of these ranges is calculated as:

1. Add the minimum values: $4 + 2 = 6$
2. Add the maximum values: $15 + 20 = 35$
3. Form the new range: $6 \leq x + y \leq 35$

Solving comparisons involving absolute values requires specific rules:

• If $|x| < a$, then $-a < x < a$.
• If $|x| > a$, then $x < -a$ or $x > a$.

Solve $|x - 5| < 7$.

Solution:

1. Original comparison: $|x - 5| < 7$
2. Equivalent double comparison: $-7 < x - 5 < 7$
3. Add 5 to each part: $-2 < x < 12$

Quadratic equations are of the form $ax^2 + bx + c = 0$, where $a \neq 0$. Solutions can be found by factoring, using the quadratic formula, or completing the square.

To solve $x^2 - 6x + 5 = 0$:

1. Factor the equation: $(x - 1)(x - 5) = 0$
2. Set each factor to zero: $x - 1 = 0$ or $x - 5 = 0$
3. Solve: $x = 1$ or $x = 5$

For equations not easily factored, use:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Example: Solve $3x^2 + 7x - 6 = 0$.

Here, $a = 3$, $b = 7$, $c = -6$. Plugging into the formula:

$x = \frac{-7 \pm \sqrt{49 + 72}}{6} = \frac{-7 \pm 11}{6}$

Solutions: $x = \frac{2}{3}$ or $x = -3$.