LET Elementary Mathematics — Elementary AlgebraStudy Notes

An algebraic expression is a combination of numbers, variables, and operations. A **term** is a single unit: the number 5, the variable x, or a product like -3xy². The **coefficient** is the numerical factor (in -3xy², the coefficient is -3). The **degree** of a term is the sum of all variable exponents (in -3xy², the degree is 2). **Like terms** have identical variable parts and can be combined by adding or subtracting their coefficients. For example, 4x and 3x are like terms, so 4x + 3x = 7x. However, 4x and 3x² are **unlike terms** and cannot be combined. Understanding this distinction prevents the most common beginner error: treating all variables as the same. When simplifying, group like terms together, then add or subtract their coefficients. For instance, simplifying 5x² + 3x - 2x² + 4x requires first collecting: (5x² - 2x²) + (3x + 4x) = 3x² + 7x. This skill is foundational because every algebraic manipulation—from solving equations to factoring—relies on correctly combining terms.

The laws of exponents provide rules for simplifying expressions involving powers. These rules are essential for algebraic manipulation and appear frequently on the LET. **Product Rule**: When multiplying powers with the same base, add the exponents. a^m · a^n = a^(m+n). For example, x³ · x⁴ = x^(3+4) = x⁷. **Quotient Rule**: When dividing powers with the same base, subtract the exponents. a^m ÷ a^n = a^(m-n). For example, x⁵ ÷ x² = x^(5-2) = x³. **Power of a Power Rule**: When raising a power to another power, multiply the exponents. (a^m)^n = a^(mn). For example, (x²)³ = x^(2·3) = x⁶. **Power of a Product Rule**: When raising a product to a power, raise each factor to that power. (ab)^n = a^n b^n. For example, (2x)³ = 2³ · x³ = 8x³. **Zero Exponent Rule**: Any nonzero number raised to the zero power equals 1. a⁰ = 1 (where a ≠ 0). For example, 7⁰ = 1. **Negative Exponent Rule**: A negative exponent means the reciprocal. a^(-n) = 1/a^n. For example, 2^(-3) = 1/2³ = 1/8. Common mistakes: Students often add exponents when using the power of a product rule (writing (2x)³ = 2x³ instead of 8x³), or forget to reverse the sign when working with negative exponents. Always apply each rule carefully and verify your answer by checking simple cases.

A **linear equation** is an equation in which the variable(s) appear only to the first power. The fundamental principle for solving any equation is **balance**: whatever operation you perform on one side of the equation, you must perform on the other side. This principle ensures the equation remains true. The general strategy for solving a linear equation in one variable is: (1) clear parentheses by distributing, (2) combine like terms on each side, (3) move all variable terms to one side and all constant terms to the other (using inverse operations), and (4) divide by the coefficient of the variable. **Inverse operations** are operations that undo each other: addition and subtraction are inverses, multiplication and division are inverses. For example, to solve 3(2x - 4) = 2x + 4, first distribute 3 to get 6x - 12 = 2x + 4. Then subtract 2x from both sides: 4x - 12 = 4. Next, add 12 to both sides: 4x = 16. Finally, divide both sides by 4: x = 4. Always **check** your solution by substituting it back into the original equation to verify. A common teaching error is skipping the check; in DepEd's problem-solving framework, verification is a critical step that builds mathematical reasoning (BEC Mathematics K–12, p. 41).

A **linear inequality** is a statement that uses the symbols <, >, ≤, or ≥ instead of =. Solving inequalities is almost identical to solving equations, with one critical exception: **when you multiply or divide both sides by a negative number, you must reverse the inequality symbol**. This rule is essential and accounts for many lost points on the LET. The reason is algebraic: the set of numbers is ordered, and multiplying by a negative reverses that order. For example, if 3 < 5, then multiplying both sides by -1 gives -3 and -5, and -3 is not less than -5; instead, -3 > -5. The solution to an inequality is typically a **range of values** (all numbers greater than 5, all numbers less than or equal to -2, etc.), often displayed on a number line. Use an open circle on a number line for < or >, and a closed circle for ≤ or ≥. For example, solving -2x < 6 gives x > -3 (remember to reverse the symbol), which on a number line is a closed circle at -3 with an arrow extending to the right. Inequalities appear in word problems about constraints, budgets, and measurements, making them practical for elementary classroom scenarios (e.g., "If each pencil costs ₱3 and you have ₱20, what is the maximum number you can buy?").

A **system of linear equations** is a set of two or more equations in two or more variables, and the solution is the ordered pair (or triple, etc.) that satisfies **all** equations simultaneously. For a system of two equations in two variables, three possible outcomes exist: **(1) One unique solution** (the lines intersect at exactly one point; the system is **consistent and independent**); (2) **No solution** (the lines are parallel and never intersect; the system is **inconsistent**); (3) **Infinitely many solutions** (the lines are identical; the system is **consistent and dependent**). Three main methods solve systems: **Substitution**, **Elimination**, and **Graphing**. **Substitution** works well when one equation can be easily solved for a variable. Solve that variable in terms of the other, then substitute the expression into the second equation. For example, given x + 2y = 11 and 3x - y = 5, solve the first for x to get x = 11 - 2y, then substitute into the second: 3(11 - 2y) - y = 5, which simplifies to 33 - 7y = 5, giving y = 4 and x = 3. **Elimination** is efficient when the coefficients align or can be scaled to match. Multiply one or both equations so that one variable has opposite coefficients in the two equations, then add them to eliminate that variable. For the system 2x + 3y = 8 and x - 2y = 1, multiply the second by 2 to get 2x - 4y = 2. Subtracting this from the first: (2x + 3y) - (2x - 4y) = 8 - 2, which gives 7y = 6, so y = 6/7, and x can be found by back-substitution. **Graphing** provides a visual check: plot both lines and identify the intersection point. Always **check** your solution in both original equations.

**Special products** are multiplication patterns that occur repeatedly in algebra. Memorizing these patterns in both directions (expanding and factored form) is essential for success on the LET. **Square of a Binomial**: (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b². Notice that the middle term is always twice the product of the two terms, and the sign of that middle term matches the sign in the binomial. For example, (x + 3)² = x² + 2(x)(3) + 9 = x² + 6x + 9, and (x - 5)² = x² - 2(x)(5) + 25 = x² - 10x + 25. **Difference of Two Squares**: (a + b)(a - b) = a² - b². This is unique because the middle terms cancel, leaving only the difference of the squares. For example, (x + 7)(x - 7) = x² - 49. **Product of Binomials (FOIL)**: (x + m)(x + n) = x² + (m + n)x + mn. The constant term is the product of m and n, and the middle coefficient is their sum. For example, (x + 3)(x + 5) = x² + 8x + 15. **Factoring** reverses multiplication. Always begin by factoring out the **greatest common monomial factor** first—this step is often overlooked but essential. For example, 6x² + 9x = 3x(2x + 3). Then apply the appropriate factoring technique: **(1) Difference of squares**: x² - 25 = (x + 5)(x - 5). **(2) Perfect-square trinomial**: x² + 6x + 9 = (x + 3)² and x² - 10x + 25 = (x - 5)². **(3) General trinomial** x² + bx + c: find two numbers whose product is c and whose sum is b. For x² - 5x + 6, the numbers -2 and -3 work (product is 6, sum is -5), so x² - 5x + 6 = (x - 2)(x - 3). Factoring is foundational for solving quadratic equations and simplifying rational expressions. A teaching note: the DepEd BEC emphasizes factoring as part of algebraic thinking (Grade 7–8), and many LET word problems hinge on correctly factoring expressions to simplify and solve.

A **quadratic equation** is an equation of the form ax² + bx + c = 0, where a ≠ 0. Three methods solve quadratic equations: **Method 1: Factoring** is the fastest if the quadratic factors neatly. Factor the left side, then apply the **zero-product property**: if a product equals zero, then at least one factor must equal zero. For example, solving x² - 5x + 6 = 0 factors as (x - 2)(x - 3) = 0, so x = 2 or x = 3. **Method 2: Square Root** works when there is no linear (bx) term. For x² = 49, take the square root of both sides: x = ±7 (note the ± symbol, indicating two solutions). More generally, (x - h)² = k gives x - h = ±√k, so x = h ± √k. **Method 3: Quadratic Formula** works for any quadratic and is the most reliable. The formula is x = [-b ± √(b² - 4ac)] / (2a), where a, b, c are the coefficients from ax² + bx + c = 0. For example, solving 2x² + 3x - 5 = 0 with a = 2, b = 3, c = -5: the discriminant is b² - 4ac = 9 - 4(2)(-5) = 9 + 40 = 49, so x = (-3 ± √49) / 4 = (-3 ± 7) / 4, giving x = 4/4 = 1 or x = -10/4 = -5/2. The **discriminant** b² - 4ac determines the nature of the roots: if positive, there are two distinct real roots; if zero, there is one repeated real root; if negative, there are no real roots (only complex roots, which are beyond elementary algebra). Understanding which method to use is a strategic skill: try factoring first (fast), use the square root method if bx is absent, and resort to the formula if the others fail. Common mistakes include forgetting the ± symbol in the square root method, making arithmetic errors in the quadratic formula, and not fully simplifying the result.

The ability to translate written sentences into algebraic equations is the highest-value skill on the LET algebra section and is critical for teaching elementary students the DepEd skill of "translating word problems to number sentences." The process requires careful reading and matching words to mathematical symbols. **Key translation phrases**: "the sum of a number and 7" → n + 7; "a number decreased by 3" → n - 3; "5 less than a number" → n - 5 (note: NOT 5 - n; order matters); "twice a number" → 2n; "the product of a number and 4" → 4n; "the quotient of a number and 2" → n/2; "a number squared" → n². **Consecutive integers** are represented as n, n + 1, n + 2, etc., and **consecutive even or odd integers** as n, n + 2, n + 4, etc. After translating and solving, always **check your answer against the original words**, not just the equation—this ensures the solution makes practical sense. **Common word-problem types on the LET**: (1) **Number problems**: defining a variable and translating phrases about relationships. Example: "The sum of three consecutive integers is 72; find the integers." Let n, n + 1, n + 2 be the integers. Then n + (n + 1) + (n + 2) = 72, giving 3n + 3 = 72, so n = 23, and the integers are 23, 24, 25. (2) **Age problems**: setting up equations based on relationships and time passages. Example: "Maria is 3 times as old as her son. In 12 years she will be twice as old as her son. Find their present ages." Let the son be x; Maria is 3x. In 12 years: 3x + 12 = 2(x + 12). Solving: 3x + 12 = 2x + 24, so x = 12 (son's age) and Maria is 36. (3) **Coin/Money problems**: two equations—one for the count of items, one for their value. Example: "A piggy bank contains ₱5 and ₱10 coins, 12 coins in all, with a total value of ₱95. How many of each?" Let x = ₱5 coins and y = ₱10 coins. Count: x + y = 12. Value: 5x + 10y = 95. Solving: from the first, x = 12 - y. Substitute: 5(12 - y) + 10y = 95, so 60 - 5y + 10y = 95, giving 5y = 35 and y = 7. Thus x = 5, and there are 5 five-peso coins and 7 ten-peso coins. (4) **Motion/Distance problems**: using distance = rate × time. Example: "Two buses leave a station traveling in opposite directions. One travels at 60 km/h, the other at 80 km/h. After how many hours will they be 280 km apart?" Let t = time in hours. Distance for first bus: 60t. Distance for second bus: 80t. Total distance: 60t + 80t = 280, so 140t = 280 and t = 2 hours. (5) **Perimeter/Area problems**: using geometric formulas. Example: "A rectangular garden's length is 3 m more than its width, and its perimeter is 26 m. Find its dimensions." Let w = width; length = w + 3. Perimeter: 2(w + w + 3) = 26, so 4w + 6 = 26, giving w = 5 m and length = 8 m. A teaching strategy per DepEd frameworks: always encourage students to define the variable clearly, list what you know and what you need to find, translate step by step, solve, and verify against the original problem statement.

**Evaluating an expression** means replacing each variable with its given value and calculating the result, following the order of operations (PEMDAS: Parentheses, Exponents, Multiplication and Division from left to right, then Addition and Subtraction from left to right). When substituting negative values, always wrap them in parentheses to avoid sign errors. For example, if x = -2 and y = 3, then 2x² - 3y = 2(-2)² - 3(3) = 2(4) - 9 = 8 - 9 = -1. **Simplifying rational expressions** (fractions containing variables, also called algebraic fractions) requires factoring the numerator and denominator, then cancelling common factors. A **rational expression** is undefined when its denominator equals zero, so always note restrictions. For example, (x² - 9)/(x - 3) factors as [(x + 3)(x - 3)] / (x - 3). Cancelling (x - 3) gives x + 3, **with the restriction that x ≠ 3**. This restriction must be stated because the original expression is undefined at x = 3, even though the simplified form appears to be defined there. Simplifying rational expressions is a reversal of factoring and appears frequently when solving rational equations on the LET. Common mistakes include cancelling across addition (e.g., incorrectly simplifying (x + 3) / (x + 2) as 3/2), which violates the algebra principle that only multiplicative factors cancel. Always factor first, then cancel only matching factors. For complex fractions (a fraction within a fraction), multiply the numerator by the reciprocal of the denominator.

A **function** is a relation where each input has exactly one output. Written f(x) = 2x - 5, the notation f(x) reads as "f of x" and represents the output when x is the input. Evaluating f(3) means substituting 3 for x: f(3) = 2(3) - 5 = 1. The **Cartesian plane** (named after René Descartes) is a coordinate system formed by two perpendicular number lines: the horizontal x-axis and the vertical y-axis, intersecting at the origin (0, 0). Any point is located by an ordered pair (x, y), where x is the horizontal distance from the origin (negative to the left, positive to the right) and y is the vertical distance (negative downward, positive upward). The plane is divided into four **quadrants**: Quadrant I (upper right, both coordinates positive), Quadrant II (upper left, x negative, y positive), Quadrant III (lower left, both negative), and Quadrant IV (lower right, x positive, y negative). A **linear function** graphs as a straight line and is often written in **slope-intercept form**: y = mx + b, where **m is the slope** and **b is the y-intercept** (the point where the line crosses the y-axis, at (0, b)). The **slope** measures the steepness of the line as "rise over run": for any two points (x₁, y₁) and (x₂, y₂) on the line, m = (y₂ - y₁) / (x₂ - x₁). A positive slope means the line rises to the right; a negative slope means it falls to the right; a zero slope is a horizontal line; and an undefined slope is a vertical line (which is not a function). Slope is the algebra behind rates of change and connects directly to the rate problems in rational numbers and measurement sections. Understanding slope is essential for teaching elementary students about proportional relationships (DepEd BEC Grade 6, Mathematics).

A **radical** is the symbol √, which asks for a root. **√a** (the square root of a) is the number that, when squared, equals a. For example, √49 = 7 because 7² = 49. Every positive number has two square roots: the positive (principal) square root and its negative opposite. We write √49 = 7 (positive root) and -√49 = -7 (negative root). **Simplifying radicals** means extracting perfect square factors from under the radical sign. For example, √50 = √(25 · 2) = √25 · √2 = 5√2. This uses the property **√(ab) = √a · √b** (the square root of a product is the product of the square roots), which is valid for non-negative a and b. To simplify √50, first find the largest perfect square that divides 50. The perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, ... Since 25 divides 50 (50 = 25 · 2), we have √50 = 5√2. A **radical equation** is an equation containing a radical, like √(x + 3) = 5. To solve, isolate the radical, square both sides (or raise to the appropriate power), solve the resulting equation, and **always check the solution in the original equation** because squaring can introduce extraneous solutions (solutions that satisfy the squared equation but not the original). For example, solving √(x + 3) = 5: square both sides to get x + 3 = 25, so x = 22. Check: √(22 + 3) = √25 = 5 ✓. Elementary algebra typically covers only square roots; cube roots and higher roots appear in more advanced courses. Radicals are foundational for the Pythagorean theorem and distance formula, both of which appear in geometry on the LET.

Success on the LET algebra section requires not only procedural skill but also strategic thinking and awareness of common errors. **Strategy 1: Identify the problem type.** Determine whether you are simplifying an expression, solving an equation, solving a system, translating a word problem, or simplifying a rational expression. Each requires a different approach. **Strategy 2: Choose the most efficient method.** For quadratics, try factoring first (fast); if that fails, use the quadratic formula. For systems, choose substitution if one equation easily isolates a variable, or elimination if coefficients align. **Strategy 3: Check your work.** Substituting the solution back into the original equation (or the original words of a word problem) is not optional—it is a final verification step that catches errors and confirms correctness. Per DepEd policy, verification is a key component of mathematical problem-solving. **Strategy 4: Watch for sign errors.** Most algebra mistakes on the LET are sign slips, not conceptual gaps. When distributing a negative (e.g., -2(x - 3) = -2x + 6), take extra care. Wrap negative values in parentheses when substituting: f(-2) not f(-2), written as f((-2)). **Strategy 5: Understand the reason behind each rule.** Knowing *why* a rule works (e.g., why the inequality reverses when multiplying by a negative) deepens understanding and prevents mechanical errors. For example, -2x < 6 becomes x > -3 because dividing both sides by -2 flips the order: -3 > -2 · (-3) / (-2), which is -3 > 3 ... wait, let me recalculate: -2(-3) = 6, so the inequality -2x < 6 means that 2|x| > 6, or |x| > 3, so x > 3 or x < -3. Actually, the correct approach is: -2x < 6, divide by -2 and reverse: x > -3. Check: x = 0 gives -2(0) < 6, which is 0 < 6 ✓, and 0 > -3 ✓. **Common Pitfalls**: (1) **Forgetting the ± in the square root method.** x² = 9 has two solutions, x = 3 and x = -3, not just x = 3. (2) **Cancelling non-factors.** (x + 3) / (x + 2) cannot be simplified to 3/2; only multiplicative factors cancel. (3) **Order in word problems.** "5 less than n" is n - 5, not 5 - n. (4) **Not checking solutions.** Always substitute back. (5) **Misapplying exponent laws.** (2x)³ ≠ 2x³; it is 8x³. (6) **Treating unlike terms as like terms.** 4x and 4x² cannot be combined. These pitfalls recur because they reflect conceptual confusion, not mere carelessness. Addressing them directly in study and practice prevents repeated errors on the exam.