4.1 Elementary Algebra & Linear Equations

4.1 Elementary Algebra & Linear Equations

Basic algebra is part of the WASSCE-level mathematics that sets the academic standard for recruit entry. In military operations, personnel frequently solve logistics and resource allocation problems where critical quantities are unknown—estimating convoy fuel consumption, balancing supply manifests, distributing ammunition crates, or determining duty rosters. Algebra replaces guesswork with a precise symbolic language of variables and equations that enables fast, manual calculations without electronic aids.

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Expressions, Brackets, and Factorization

An algebraic expression combines numbers, operations, and variables representing unknown quantities.

Terms and Collecting Like Terms

• Terms are components separated by $+$ or $-$ signs ($5x^2$, $-3xy$, $+7$).
• Coefficients are multipliers preceding variables ($5$, $-3$); standalone numbers are constants ($+7$).
• Like terms share identical variable letters and exponents. Only like terms can be combined:

$$7a - 3b + 4 - 2a + 8b - 9 = 5a + 5b - 5$$

Expanding Brackets

Distribute the outside factor across enclosed terms ($a(b + c) = ab + ac$). Negative factors reverse all signs:

$$-4(2x - 3y + 5) = -8x + 12y - 20$$

For binomials $(a + b)(c + d)$, apply FOIL:

$$(2x + 5)(x - 3) = 2x^2 - 6x + 5x - 15 = 2x^2 - x - 15$$

Factorization Methods

Factorization rewrites an expression as a product of simpler factors:

1. Common Monomial Factor: Extract the greatest common divisor:

$$12x^2y - 18xy^2 = 6xy(2x - 3y)$$

1. Grouping: Group four terms in pairs to find a shared binomial factor:

$$3ax - 6ay + bx - 2by = 3a(x - 2y) + b(x - 2y) = (3a + b)(x - 2y)$$

1. Difference of Two Squares: Apply $a^2 - b^2 = (a - b)(a + b)$:

$$25x^2 - 49 = (5x - 7)(5x + 7)$$

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Solving Single-Variable Linear Equations

A linear equation in one variable takes the standard form $ax + b = c$ ($a \neq 0$). The objective is isolating the variable through inverse balancing operations.

Clearing Denominators with the LCM

When equations contain fractions, multiply all terms by the Lowest Common Multiple (LCM) of the denominators:

$$\frac{2(x - 1)}{3} - \frac{x + 2}{4} = 3$$

1. Multiply every term by the LCM ($12$):

$$4[2(x - 1)] - 3(x + 2) = 36$$

1. Expand brackets and collect terms:

$$8x - 8 - 3x - 6 = 36 \implies 5x - 14 = 36$$

1. Isolate $x$:

$$5x = 50 \implies x = 10$$

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Systems of Simultaneous Linear Equations

A simultaneous system consists of two linear equations with two unknown variables satisfied concurrently:

$$a_1x + b_1y = c_1 \quad \text{and} \quad a_2x + b_2y = c_2$$

Elimination Method

Multiply equations by integer factors to match the absolute coefficients of one variable. Add equations if signs differ; subtract if signs match:

• For $2x + 3y = 19$ and $3x - y = 12$, multiply the second equation by $3$: $9x - 3y = 36$.
• Adding the equations eliminates $y$: $11x = 55 \implies x = 5$.
• Substitute $x = 5$: $3(5) - y = 12 \implies y = 3$.

Substitution Method

Isolate a variable with coefficient $\pm 1$ and substitute into the other equation:

• From $x + 2y = 11$, isolate $x = 11 - 2y$.
• Substitute into $3x - 4y = 3$:

$$3(11 - 2y) - 4y = 3 \implies 33 - 10y = 3 \implies y = 3$$

• Then $x = 11 - 2(3) = 5$.

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Word-to-Algebra Translation: Military Logistics Models

Translating situational word problems into equations is a core aptitude competency.

Logistics and Price-Quantity Systems

If $4$ ammunition crates and $3$ supply boxes weigh $130\text{ kg}$, while $2$ crates and $5$ boxes weigh $170\text{ kg}$, define crate weight $c$ and box weight $b$:

$$4c + 3b = 130 \quad \text{and} \quad 2c + 5b = 170$$

Age Problems

Anchor time shifts to current baseline ages:

• Current ages: Officer $A$, Recruit $B$.
• "Four years ago": $(A - 4)$ and $(B - 4)$. If the officer was four times as old: $A - 4 = 4(B - 4)$.
• "In six years": $(A + 6)$ and $(B + 6)$.

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Linear Inequalities and Number Line Representations

A linear inequality models allowable operational boundaries using relational symbols: $<$, $\le$, $>$, $\ge$.

The Inversion Rule

Transformations mirror equations, with one vital rule:

Solve $-4x + 6 \le 26$:

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

Graphing on a Number Line

• Open circle ($\circ$): Strict inequality ($<$ or $>$); boundary value excluded.
• Closed circle ($\bullet$): Inclusive inequality ($\le$ or $\ge$); boundary value included.
• Ray direction: Points right for greater values ($>$ or $\ge$); points left for smaller values ($<$ or $\le$).
• Bounded interval: $-3 \le x < 4$ features a solid dot at $-3$, an open circle at $4$, and shading between them.

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Worked Examples with Step-by-Step Solutions

Example 1: Logistics Simultaneous System

Problem: A logistics depot dispatches two deliveries. Truck 1 carries $30$ boot pairs and $40$ uniforms weighing $110\text{ kg}$. Truck 2 carries $20$ boot pairs and $50$ uniforms weighing $120\text{ kg}$. Find the unit weight of boots ($b$) and uniforms ($u$).

Solution:

1. Divide both equations by $10$: $3b + 4u = 11$ and $2b + 5u = 12$.
2. Multiply by $2$ and $3$ to match $b$: $6b + 8u = 22$ and $6b + 15u = 36$.
3. Subtract equations: $7u = 14 \implies u = 2\text{ kg}$.
4. Substitute $u = 2$: $3b + 4(2) = 11 \implies 3b = 3 \implies b = 1\text{ kg}$. Final Answer: Boots weigh $1\text{ kg}$; uniforms weigh $2\text{ kg}$.

Example 2: Ration Distribution Problem

Problem: A sergeant distributes $96$ ration bars between Squad A and Squad B. Squad A has $4$ more soldiers than Squad B. If each soldier receives $3$ bars, how many soldiers are in Squad B?

Solution:

1. Let Squad B have $s$ soldiers; Squad A has $s + 4$. Total soldiers: $2s + 4$.
2. Form the equation: $3(2s + 4) = 96$.
3. Divide by $3$: $2s + 4 = 32 \implies 2s = 28 \implies s = 14$. Final Answer: Squad B has $14$ soldiers.