LET Elementary Mathematics — Elementary AlgebraSummary

An algebraic expression combines numbers (constants), variables, and operations. A term is a single number or a product of numbers and variables (e.g., 5, x, −3xy²). Like terms have identical variable parts and can be combined by adding or subtracting their coefficients: 4x + 3x = 7x, but 4x + 3x² cannot be combined. The coefficient is the numerical factor (in −5x², the coefficient is −5), and the degree of a term is the sum of all variable exponents (in 3x²y, the degree is 3). Simplifying expressions means combining like terms and using the distributive property to remove parentheses: 3(2x − 4) + 5x = 6x − 12 + 5x = 11x − 12.

Exponents follow consistent rules that allow us to simplify products, quotients, and powers of powers without expanding everything out. The product rule: a^m · a^n = a^(m+n) (add exponents when multiplying like bases). The quotient rule: a^m ÷ a^n = a^(m−n) (subtract exponents when dividing like bases). Power of a power: (a^m)^n = a^(mn) (multiply exponents). Power of a product: (ab)^n = a^n · b^n (distribute the exponent). Zero exponent: a^0 = 1 for any a ≠ 0. Negative exponent: a^(−n) = 1/a^n (a negative exponent represents a reciprocal). For example, (2x³)² · 3x = 4x⁶ · 3x = 12x⁷, because we square 2 to get 4 and multiply 3 by 2 to get 6, then add exponents (6 + 1) when multiplying.

A linear equation has the variable to the first power only, such as 3x + 5 = 14. To solve, use inverse operations to isolate the variable while keeping both sides balanced. The general strategy: (1) distribute to clear parentheses, (2) combine like terms on each side, (3) move variable terms to one side and constants to the other, (4) divide by the coefficient of the variable. For example, to solve 3(2x − 4) = 2x + 4: distribute to get 6x − 12 = 2x + 4; subtract 2x from both sides to get 4x − 12 = 4; add 12 to both sides to get 4x = 16; divide by 4 to get x = 4. Always check by substituting back: 3(2·4 − 4) = 3(4) = 12 and 2·4 + 4 = 12 ✓.

Linear inequalities (<, >, ≤, ≥) are solved almost identically to equations, with one crucial exception: when you multiply or divide both sides by a negative number, you must reverse the inequality symbol. For example, to solve −2x < 6: divide both sides by −2, and reverse < to > to get x > −3. Solutions are ranges, not single values, and are often represented on a number line or in interval notation. For x > −3, every number greater than −3 is a solution. Checking: x = −2 gives −2(−2) < 6, or 4 < 6 ✓; x = −4 gives −2(−4) < 6, or 8 < 6 ✗, confirming that −4 is not a solution.

A system of two equations in two variables (x and y) requires finding an ordered pair (x, y) that satisfies both equations simultaneously. Three main solution methods exist: (1) Substitution: solve one equation for a variable, substitute that expression into the other equation, and solve for the remaining variable. Example: for x + 2y = 11 and 3x − y = 5, from the first x = 11 − 2y; substitute into the second to get 3(11 − 2y) − y = 5, simplify to 33 − 7y = 5, so y = 4 and x = 3. Solution: (3, 4). (2) Elimination: multiply one or both equations by constants so that adding or subtracting them cancels one variable. (3) Graphing: plot both lines on the Cartesian plane; their intersection point is the solution. A system with one solution is consistent and independent; parallel lines (no solution) are inconsistent; coincident lines (infinitely many solutions) are dependent.

Special products are patterns of multiplication that occur frequently and should be memorized and recognized instantly, both in expanded and factored form. Square of a binomial: (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b² (note the middle term is twice the product of a and b). Difference of two squares: (a + b)(a − b) = a² − b² (no middle term; very useful for factoring). General product of binomials (FOIL): (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15. Factoring reverses these patterns. Systematic factoring strategy: (1) Always factor out any common monomial factor first (e.g., 6x² + 9x = 3x(2x + 3)). (2) Recognize difference of squares: x² − 25 = (x + 5)(x − 5). (3) Recognize perfect-square trinomial: x² + 6x + 9 = (x + 3)². (4) For a general trinomial x² + bx + c, find two numbers whose product is c and sum is b. For x² − 5x + 6, those numbers are −2 and −3, giving (x − 2)(x − 3). For trinomials with a leading coefficient ≠ 1 (e.g., 2x² + 5x + 3), use the grouping method or trial factors.

A quadratic equation has the form ax² + bx + c = 0 (a ≠ 0) and represents a second-degree polynomial. Three reliable solution methods exist: (1) Factoring (when it factors neatly): set the equation equal to zero, factor completely, and apply the zero-product property (if a product equals zero, at least one factor must be zero). Example: x² − 5x + 6 = 0 factors as (x − 2)(x − 3) = 0, so x = 2 or x = 3. (2) Square-root method (when there is no linear term, bx = 0): isolate x² and take the square root of both sides, remembering the ± symbol. Example: x² = 49 gives x = ±7. (3) Quadratic formula (always works): x = [−b ± √(b² − 4ac)] / (2a). For 2x² + 3x − 5 = 0, a = 2, b = 3, c = −5; discriminant = 9 − 4(2)(−5) = 49; x = (−3 ± 7) / 4, so x = 1 or x = −5/2. The discriminant b² − 4ac determines the nature of the roots: positive = two real roots, zero = one repeated real root, negative = no real roots (complex roots).

This is the most tested algebra skill on the LET: the ability to read a word problem and write an equation that models the situation. Common translation phrases: 'the sum of a number and 7' = n + 7; '5 less than a number' = n − 5 (note order: it's n − 5, not 5 − n); 'twice a number' = 2n; 'the product of a number and 4' = 4n; 'the quotient of a number and 2' = n/2. For sequences: consecutive integers = n, n + 1, n + 2; consecutive even integers = n, n + 2, n + 4; consecutive odd integers = n, n + 2, n + 4. Example: 'The sum of three consecutive integers is 72.' Translate: n + (n + 1) + (n + 2) = 72, simplify 3n + 3 = 72, solve n = 23. The integers are 23, 24, 25. Always verify against the original words: 23 + 24 + 25 = 72 ✓. Common problem types: number problems (simple relationships between quantities), coin/money problems (count and value), age problems (relationships between people's ages at different times), motion problems (distance = rate × time), and geometry problems (perimeter, area).

Evaluating an expression means substituting each variable with its given value and applying the order of operations. Example: if x = 3 and y = −2, then 2x² − 3y = 2(3)² − 3(−2) = 2(9) + 6 = 18 + 6 = 24. A key error-prevention strategy is to wrap negative values in parentheses: write 2(−2)² not 2−2², since (−2)² = 4 but −2² = −4. Simplifying rational expressions (algebraic fractions) means factoring the numerator and denominator, then cancelling common factors. Example: (x² − 9)/(x − 3). Factor the numerator as a difference of squares: (x + 3)(x − 3). Then (x² − 9)/(x − 3) = [(x + 3)(x − 3)] / (x − 3) = x + 3 (for x ≠ 3). The restriction x ≠ 3 must be noted because dividing by zero is undefined. This simplification is factoring and cancelling in reverse; it appears frequently on the LET embedded in larger problems.

A function assigns exactly one output (y) to each input (x); it is written as f(x) or y = f(x). For example, f(x) = 2x² − 3x + 1; to find f(−2), substitute x = −2: f(−2) = 2(4) + 6 + 1 = 15. The Cartesian plane locates points using ordered pairs (x, y): x is the horizontal (left-right) coordinate, y is the vertical (up-down) coordinate, and the axes meet at the origin (0, 0). The plane is divided into four quadrants (I, II, III, IV), numbered counter-clockwise starting from the upper right. A linear function graphs as a straight line and is 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)). Slope measures steepness as rise/run: m = (y₂ − y₁) / (x₂ − x₁) for any two points (x₁, y₁) and (x₂, y₂) on the line. A positive slope rises to the right, a negative slope falls to the right, zero slope is horizontal, and undefined slope is vertical. Slope is the rate of change in a linear relationship, directly linking to rate problems in applied contexts.

A radical (√) symbol asks for a root. The square root √a asks: what number squared equals a? For example, √49 = 7 because 7² = 49. To simplify a radical, factor the radicand (the number under the radical) and extract perfect square factors. Example: √50 = √(25 · 2) = √25 · √2 = 5√2. The rule is √(ab) = √a · √b (the square root of a product equals the product of square roots). A radical is simplified when no perfect-square factors remain under the radical (other than 1). Rationalizing the denominator means rewriting a fraction with a radical in the denominator so that there is no radical in the denominator. For example, 1/√2 = 1/√2 · √2/√2 = √2/2. This technique is useful for exact answers.

The distributive property is the conceptual core connecting all algebra topics. It allows you to multiply out (expand) expressions like 3(2x − 4) = 6x − 12, and it also underpins factoring (the reverse: 6x − 12 = 3(2x − 4)). When you solve equations, you use the distributive property to clear parentheses. When you factor quadratics, you apply it implicitly. When you simplify rational expressions, you factor using it. Teaching this property as the 'glue' holding algebra together helps pupils see relationships rather than memorizing isolated techniques.

Understanding exponent laws allows you to simplify any polynomial expression, from (2x³)² · 3x to x⁵/x² to (3a²b)³. These laws also appear when solving quadratic equations (e.g., recognizing that (x + 2)² = x² + 4x + 4 is a special product that uses exponent concepts). Without fluency with exponent laws, many algebraic manipulations feel arbitrary; with them, patterns emerge.

If you truly understand (a + b)² = a² + 2ab + b², you can recognize its factored form instantly: x² + 6x + 9 is (x + 3)² because 6x is twice 3x. Similarly, understanding (a + b)(a − b) = a² − b² lets you factor x² − 25 as (x + 5)(x − 5) without guessing. The more fluent you are with special products in the forward direction, the faster you recognize them backward during factoring.

The zero-product property (if ab = 0 then a = 0 or b = 0) is the bridge between factoring and quadratic equations. You cannot solve 2x² + 5x + 3 = 0 by factoring unless you can first factor 2x² + 5x + 3 as (2x + 3)(x + 1). Conversely, if you cannot factor, the quadratic formula becomes your reliable backup. Understanding this relationship helps pupils see that factoring is valuable (it is faster) but not always necessary (the formula always works).

A single word problem may require you to translate (defining a variable), simplify (combining like terms), solve (linear or quadratic equation), and verify (checking the solution against the original context). Consecutive-integer problems require translation and solving; age problems require two equations (a system); motion problems require equations with fractions. Word problems are the 'capstone' that ties all prior skills together and demonstrate that algebra models real situations.

A system of two equations in two unknowns models any situation with two constraints. Coin problems (count constraint and value constraint), age problems (present-age relationship and future-age relationship), and mixture problems all naturally lead to systems. Understanding substitution and elimination methods gives pupils tools to solve these real-world scenarios, reinforcing that algebra is not abstract but a powerful language for describing and solving problems.

A linear function f(x) = mx + b is both an algebraic expression and a geometric object (a line on the Cartesian plane). The slope m is both the coefficient of x in the equation and the rise/run of the line. Understanding this connection means pupils can translate between algebraic and graphical representations, deepening conceptual understanding. When solving a system graphically, the intersection point of two lines (a visual, geometric event) corresponds to the algebraic solution.

Solving an inequality is almost identical to solving an equation, except the answer is a range (or set) rather than a single value. This is a natural extension and helps pupils see that algebra can describe sets of solutions, not just individual answers. The critical rule (reverse the inequality when multiplying/dividing by a negative) is a unique feature that distinguishes inequalities from equations and requires careful attention.

A classic LET item: 'A cashier has 15 coins (₱5 and ₱10), totaling ₱110. How many of each coin?' This translates to a system of two equations: x + y = 15 (count) and 5x + 10y = 110 (value). Solving by substitution or elimination gives the answer. Teaching pupils to set up such problems models real scenarios they will encounter and shows that algebra solves practical problems. This connects to financial literacy in the K–12 BEC.

A parent and child problem: 'A mother is 3 times as old as her son. In 12 years she will be twice as old. Find their current ages.' Let the son's age be x; the mother's is 3x. In 12 years: 3x + 12 = 2(x + 12). Solving gives x = 12 (son) and 36 (mother). These problems reinforce that algebraic relationships hold at different times and teach pupils to think about change over time—a key concept in science and social studies.

A rectangular garden has length 3 meters more than width, and perimeter 26 meters. Let width = w; then 2(w + w + 3) = 26, which simplifies to 4w + 6 = 26. Solving gives w = 5 m and length = 8 m. These problems integrate algebra and geometry, showing pupils that algebra is a tool for solving spatial problems. This directly supports the Geometry strand of the K–12 BEC.

Two vehicles travel toward each other; one goes 50 km/h and the other 60 km/h. They start 220 km apart. How long until they meet? Let t = time (hours). Distance covered: 50t + 60t = 220, so 110t = 220, giving t = 2 hours. This problem uses the relationship distance = rate × time and shows algebra modeling a dynamic scenario. It connects to physics and is practical for understanding real-world motion.

A store buys items for ₱300 each and sells for ₱450. If overhead is ₱2000 per month, how many items must be sold to break even? Let n = items sold. Revenue: 450n. Cost: 300n + 2000. Break-even: 450n = 300n + 2000, so 150n = 2000, giving n ≈ 13.33, meaning 14 items. This applies algebra to economic reasoning and supports curriculum on financial decision-making.

How much 20% sugar solution should be mixed with 50% solution to make 10 liters of 30% solution? Let x = liters of 20% solution. Then (10 − x) = liters of 50% solution. Sugar content: 0.20x + 0.50(10 − x) = 0.30(10). Solving: 0.20x + 5 − 0.50x = 3, so −0.30x = −2, giving x ≈ 6.67 liters. This problem integrates percent, ratios, and linear equations—skills tested on the LET and applicable to chemistry and health science contexts.

Worker A can complete a task in 6 hours; Worker B in 4 hours. How long if they work together? Let t = time (hours). Rate of A: 1/6 per hour. Rate of B: 1/4 per hour. Together: (1/6)t + (1/4)t = 1 (one complete task). So (5/12)t = 1, giving t = 12/5 = 2.4 hours. These problems teach reciprocals, fractions, and algebraic reasoning about rates—skills applicable to project management and workforce planning.

A farmer wants to enclose a rectangular field with 100 meters of fencing. If one side is against a river (so it needs no fence), what dimensions maximize area? Let x = width (perpendicular to river). Then length = 100 − 2x. Area A = x(100 − 2x) = 100x − 2x². This is a quadratic function; its maximum occurs at x = 25 (using calculus or vertex formula). This shows algebra and quadratics solving real optimization problems—a precursor to higher mathematics and practical planning.

The formula C = (5/9)(F − 32) converts Fahrenheit to Celsius. Evaluating: if F = 98.6°, then C = (5/9)(66.6) ≈ 37°. Solving for F given C: F = (9/5)C + 32. These are linear relationships in disguise; teaching pupils to manipulate formulas algebraically (solving for a different variable) builds flexibility and shows that algebra extends to science contexts.

A map scale is 1:50,000 (1 cm on the map = 50,000 cm in reality). If two cities are 8 cm apart on the map, what is the actual distance? Set up: 1/50,000 = 8/x, so x = 400,000 cm = 4 km. This uses proportional reasoning and linear equations to solve geometric problems—skills applicable to geography, engineering, and design.