4.2 Basic Algebra, Linear Equations & Exponents

Algebraic Expressions, Factoring & Quadratic Equations

Algebra forms the structural foundation of technical calculation in the PAF Airman selection process. Candidates must master algebraic expansion, factorization, linear systems, and exponent laws.

Standard Algebraic Identities

Polynomial expressions can often be simplified using classic identity expansions:

$$\begin{aligned} (a + b)^2 &= a^2 + 2ab + b^2 \\ (a - b)^2 &= a^2 - 2ab + b^2 \\ a^2 - b^2 &= (a + b)(a - b) \\ (a + b)^3 &= a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a + b) \\ a^3 + b^3 &= (a + b)(a^2 - ab + b^2) \\ a^3 - b^3 &= (a - b)(a^2 + ab + b^2) \end{aligned}$$

Worked Example: Factoring Quadratic Expressions

Problem: Solve the quadratic equation $2x^2 - 7x + 6 = 0$ by factoring.

Solution:

1. Find two numbers that multiply to give $a \cdot c = 2 \times 6 = 12$ and add up to $b = -7$.
2. The two numbers are $-3$ and $-4$ (since $(-3) \times (-4) = 12$ and $(-3) + (-4) = -7$).
3. Rewrite the middle term: $2x^2 - 4x - 3x + 6 = 0$.
4. Group terms: $2x(x - 2) - 3(x - 2) = 0 \implies (2x - 3)(x - 2) = 0$.
5. Set each factor to zero:
   • $2x - 3 = 0 \implies x = \frac{3}{2} = 1.5$
   • $x - 2 = 0 \implies x = 2$

Roots: $x = 1.5$ or $x = 2$.

The Quadratic Formula & Discriminant

For any quadratic equation $ax^2 + bx + c = 0$ ($a \neq 0$):

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

The term under the square root, $D = b^2 - 4ac$, is called the Discriminant:

• $D > 0$: Two distinct real roots.
• $D = 0$: Two equal real roots (one repeated root).
• $D < 0$: No real roots (two complex conjugate roots).

Linear Systems & Laws of Exponents

Systems of Simultaneous Linear Equations

A system of two linear equations in two variables takes the form:

$$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$$

Methods of Solution

1. Elimination Method: Multiply one or both equations by constants so that coefficients of one variable match, then add or subtract equations.
2. Substitution Method: Express one variable in terms of the other from one equation and substitute it into the second equation.
3. Cramer's Rule (Determinants): $$x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}$$ where $D = a_1 b_2 - a_2 b_1$, $D_x = c_1 b_2 - c_2 b_1$, and $D_y = a_1 c_2 - a_2 c_1$. (Valid when $D \neq 0$).

Worked Example: System of Linear Equations

Problem: Solve the system:

$$\begin{cases} 3x + 4y = 18 \\ 5x - 2y = 4 \end{cases}$$

Solution (Elimination Method):

• Multiply the second equation by $2$ to match the coefficient of $y$: $$2 \times (5x - 2y = 4) \implies 10x - 4y = 8$$
• Add this modified equation to the first equation: $$(3x + 4y) + (10x - 4y) = 18 + 8$$ $$13x = 26 \implies x = 2$$
• Substitute $x = 2$ into $3x + 4y = 18$: $$3(2) + 4y = 18 \implies 6 + 4y = 18 \implies 4y = 12 \implies y = 3$$

Solution Pair: $(x, y) = (2, 3)$.

Laws of Exponents and Indices

Exponent questions in PAF Airman examinations test algebraic simplification under exponential rules.

Worked Exponential Equation

Problem: Solve for $x$ in the equation: $8^{x-1} = 32^{x-3}$.

Solution:

• Express both sides with a common base of $2$: $$8 = 2^3 \implies 8^{x-1} = (2^3)^{x-1} = 2^{3(x-1)} = 2^{3x - 3}$$ $$32 = 2^5 \implies 32^{x-3} = (2^5)^{x-3} = 2^{5(x-3)} = 2^{5x - 15}$$
• Equate exponents: $$3x - 3 = 5x - 15$$ $$-3 + 15 = 5x - 3x \implies 12 = 2x \implies x = 6$$

Result: $x = 6$.