6.2 High School Algebra: Linear Equations, Inequalities & Polynomials

6.2 High School Algebra: Linear Equations, Inequalities & Polynomials

Algebraic reasoning forms the core of high school mathematics and constitutes a major portion of the AFCT Mathematics Knowledge subtest. Mastering linear equations, inequalities, polynomial manipulations, and quadratic solutions provides the essential toolkit for solving complex mathematical word problems.

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Single-Variable & Multi-Step Linear Equations

A linear equation in one variable is an equation that can be written in the form $ax + b = 0$, where $a \neq 0$. The objective when solving linear equations is to isolate the variable on one side of the equal sign.

Systematic 4-Step Solution Method

1. Clear Parentheses: Apply the distributive property $a(b + c) = ab + ac$.
2. Clear Fractions: Multiply every term on both sides by the least common denominator (LCD).
3. Combine Like Terms: Group variable terms on one side and constant terms on the opposite side.
4. Isolate Variable: Divide or multiply by the variable's coefficient.

Literal Equations (Formula Rearrangement)

AFCT questions frequently ask you to solve a geometry or physics formula for a specific variable.

Example: Solve the trapezoid area formula $A = \frac{1}{2}h(b_1 + b_2)$ for $b_1$.

• Multiply both sides by $2$: $2A = h(b_1 + b_2)$.
• Divide by $h$: $\frac{2A}{h} = b_1 + b_2$.
• Subtract $b_2$: $b_1 = \frac{2A}{h} - b_2$.

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Systems of Linear Equations

A system of linear equations consists of two or more equations with two or more variables. The solution is the ordered pair $(x, y)$ that satisfies both equations simultaneously.

Solution Methods

1. Substitution Method: Solves one equation for one variable and substitutes that expression into the second equation. Best when one coefficient is $1$ or $-1$.
2. Elimination (Addition) Method: Multiplies one or both equations by constants so that adding the equations eliminates one variable. Best for general equations.

Types of Linear Systems

Worked Example: Elimination Method

Solve the system: $\begin{cases} 3x + 4y = 18 \quad \text{--- (Eq. 1)} \\ 2x - 3y = -5 \quad \text{--- (Eq. 2)} \end{cases}$

• Multiply Eq. 1 by $3$: $9x + 12y = 54$.
• Multiply Eq. 2 by $4$: $8x - 12y = -20$.
• Add the equations to eliminate $y$: $(9x + 8x) + (12y - 12y) = 54 - 20 \implies 17x = 34 \implies x = 2$
• Substitute $x = 2$ into Eq. 1: $3(2) + 4y = 18 \implies 6 + 4y = 18 \implies 4y = 12 \implies y = 3$
• Solution: $(2, 3)$.

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Linear Inequalities & Absolute Value

Linear inequalities use order symbols: $<, \le, >, \ge$. Solving inequalities follows the same rules as equations, with one vital exception.

GOLDEN RULE OF INEQUALITIES: When multiplying or dividing both sides of an inequality by a negative number, you MUST reverse the inequality direction ($<$ becomes $>$, $\le$ becomes $\ge$, etc.).

Example: Negative Division

$-4x + 7 \ge 27 \implies -4x \ge 20 \implies \frac{-4x}{-4} \le \frac{20}{-4} \implies x \le -5$

Absolute Value Equations & Inequalities

Absolute value $|x|$ represents distance from zero on a number line (always non-negative).

• Equation: $|x| = k \implies x = k \quad \text{or} \quad x = -k \quad (k \ge 0)$.
• "Less Than" Inequality (AND): $|x| < k \implies -k < x < k$.
• "Greater Than" Inequality (OR): $|x| > k \implies x < -k \quad \text{or} \quad x > k$.

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Polynomial Operations & Factoring

A polynomial is an expression consisting of variables and coefficients combined using addition, subtraction, and non-negative integer exponents.

Polynomial Operations

• Multiplying Binomials (FOIL): $(a + b)(c + d) = \text{First } (ac) + \text{Outer } (ad) + \text{Inner } (bc) + \text{Last } (bd)$. $(2x - 5)(3x + 4) = 6x^2 + 8x - 15x - 20 = 6x^2 - 7x - 20$

Core Factoring Techniques

1. Greatest Common Factor (GCF): Factor out the highest common factor from all terms. $12x^3 - 18x^2 = 6x^2(2x - 3)$
2. Difference of Two Squares: $a^2 - b^2 = (a - b)(a + b)$. $16x^2 - 49 = (4x - 7)(4x + 7)$
3. Perfect Square Trinomials: $a^2 \pm 2ab + b^2 = (a \pm b)^2$. $x^2 - 10x + 25 = (x - 5)^2$
4. Trinomial Factoring ($x^2 + bx + c$): Find two numbers that multiply to $c$ and add to $b$. $x^2 - 7x + 12 = (x - 3)(x - 4) \quad \text{(since } -3 \times -4 = 12 \text{ and } -3 + -4 = -7$

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Quadratic Equations & The Quadratic Formula

A quadratic equation is a second-degree polynomial equation: $ax^2 + bx + c = 0 \quad (a \neq 0)$.

The Quadratic Formula

For any quadratic equation $ax^2 + bx + c = 0$, the roots are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The Discriminant ($D = b^2 - 4ac$)

The expression under the square root, $D = b^2 - 4ac$, reveals the number and nature of roots:

• If $D > 0$ and a perfect square: Two distinct rational real roots.
• If $D > 0$ and not a perfect square: Two distinct irrational real roots.
• If $D = 0$: Exactly one repeated rational real root ($x = -b / 2a$).
• If $D < 0$: No real roots (two complex conjugate roots).

Worked Example: Quadratic Solution

Solve $2x^2 - 5x - 3 = 0$.

• Identify coefficients: $a = 2$, $b = -5$, $c = -3$.
• Calculate discriminant: $D = (-5)^2 - 4(2)(-3) = 25 + 24 = 49$.
• Apply Quadratic Formula: $x = \frac{-(-5) \pm \sqrt{49}}{2(2)} = \frac{5 \pm 7}{4}$
• Solution 1: $x = \frac{5 + 7}{4} = \frac{12}{4} = 3$.
• Solution 2: $x = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2}$.
• Solution set: ${ -\frac{1}{2}, 3 }$.