Variables and algebraic expressions

LVL: FREE

MODULE: Equations and Inequalities

[EXEC: MICRO_CORE]

βœ–οΈ 1. Definition of variables and constants in algebraic contexts

πŸ“¦ Variables and Constants

  • A variable is a letter that stands for an unknown or changing number.
  • A constant is a fixed number that never changes.
  • Variables can take different values in different situations.
  • We usually use letters like xx, yy, nn, or tt for variables.
  • Constants are just regular numbers like 5, -3, or 0.5.

If x=2x = 2, then 3x3x means 3Γ—2=63 \times 2 = 6. Here xx is the variable and 3 is the constant.

πŸ’‘ Think: Variables are empty boxes waiting for numbers; constants are locked numbers.

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1. Definition of variables and constants in algebraic contexts

Definition of Variables and Constants

A variable is a symbol (commonly a letter such as xx, yy, or nn) that represents an unknown or changing quantity. A constant is a fixed numerical value that does not change within a given context.

Intuition: Variables act as placeholders that can take on different values, while constants remain the same throughout a problem.

Core Rules:

  • Variables can represent any number from a specified set (e.g., real numbers, integers).
  • Constants are specific numbers like 33, βˆ’5-5, or Ο€\pi.
  • In expressions like 3x+73x + 7, the variable is xx and the constants are 33 and 77.
  • By convention, letters near the end of the alphabet (xx, yy, zz) typically denote variables, while early letters (aa, bb, cc) may denote constants or parameters.

Consequence: Distinguishing variables from constants is essential for manipulating and solving algebraic expressions and equations.

Example: In 5nβˆ’25n - 2, nn is the variable and 55, βˆ’2-2 are constants.

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Given the algebraic expression 8y+48y + 4, which symbol represents the variable?

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βœ–οΈ 2. Translating simple linguistic phrases into algebraic expressions

πŸ—£οΈ Words to Algebra

  • "Sum" or "more than" means add: x+5x + 5.
  • "Difference" or "less than" means subtract: xβˆ’3x - 3.
  • "Product" or "times" means multiply: 4x4x.
  • "Quotient" or "divided by" means divide: x2\frac{x}{2}.
  • Order matters for subtraction and division.

"Five less than a number" translates to xβˆ’5x - 5, NOT 5βˆ’x5 - x.

πŸ’‘ Trick: Read carefullyβ€”"less than" flips the order!

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2. Translating simple linguistic phrases into algebraic expressions

Translating Linguistic Phrases into Algebraic Expressions

Translation involves converting verbal descriptions into symbolic algebraic form using variables and operations. Each phrase corresponds to a specific mathematical operation.

Intuition: Words like "sum," "difference," "product," and "quotient" directly map to addition, subtraction, multiplication, and division.

Core Rules:

  • "The sum of xx and 44" translates to x+4x + 4.
  • "Three times a number nn" translates to 3n3n.
  • "Five less than yy" translates to yβˆ’5y - 5 (order matters for subtraction).
  • "The quotient of aa and bb" translates to ab\frac{a}{b} or a/ba/b.
  • Order is critical: "xx minus 22" is xβˆ’2x - 2, not 2βˆ’x2 - x.

Consequence: Accurate translation is foundational for setting up equations and solving word problems.

Example: "Twice a number increased by seven" translates to 2x+72x + 7.

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MOD: TRANSLATE

Translate the following phrase into an algebraic expression: 'The sum of four times a number kk and nine'.

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βœ–οΈ 3. Evaluating expressions by substituting specific numerical values

πŸ”’ Plug and Calculate

  • Substitution means replacing the variable with a given number.
  • Write the expression, then replace every variable with its value.
  • Follow order of operations: parentheses, exponents, multiply/divide, add/subtract.
  • Always do multiplication before addition unless parentheses say otherwise.

Evaluate 2x+32x + 3 when x=4x = 4: Replace to get 2(4)+3=8+3=112(4) + 3 = 8 + 3 = 11.

πŸ’‘ Remember: Substitute first, then calculate step-by-step.

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3. Evaluating expressions by substituting specific numerical values

Evaluating Expressions by Substitution

Evaluation means replacing each variable in an algebraic expression with a given numerical value and computing the result using order of operations.

Intuition: Substitution transforms an abstract expression into a concrete number by assigning specific values to variables.

Core Rules:

  • Replace every occurrence of the variable with the given number.
  • Use parentheses when substituting to avoid sign errors (e.g., substituting βˆ’3-3 for xx in x2x^2 gives (βˆ’3)2=9(-3)^2 = 9).
  • Follow the order of operations: parentheses, exponents, multiplication/division (left to right), addition/subtraction (left to right).
  • Sign precision: Substituting negative values requires careful handling of operations.

Consequence: Evaluation allows us to test expressions with specific inputs and verify formulas.

Example: Evaluate 3xβˆ’53x - 5 when x=4x = 4: substitute to get 3(4)βˆ’5=12βˆ’5=73(4) - 5 = 12 - 5 = 7.

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Evaluate the expression 5x+85x + 8 when x=6x = 6.

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βœ–οΈ 4. Distinguishing between an algebraic expression and an equation

βš–οΈ Expression vs Equation

  • An expression is a math phrase with no equals sign: 3x+73x + 7.
  • An equation has an equals sign connecting two expressions: 3x+7=223x + 7 = 22.
  • Expressions can be simplified or evaluated.
  • Equations can be solved to find the variable's value.
  • Think of an equation as a balance scale.

5xβˆ’25x - 2 is an expression; 5xβˆ’2=135x - 2 = 13 is an equation.

πŸ’‘ Key: No equals sign = expression; equals sign = equation.

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4. Distinguishing between an algebraic expression and an equation

Distinguishing Expressions from Equations

An algebraic expression is a combination of variables, constants, and operations without an equality sign. An equation is a statement asserting that two expressions are equal, connected by an equals sign.

Intuition: Expressions represent quantities; equations represent relationships or conditions that variables must satisfy.

Core Rules:

  • Expressions can be simplified or evaluated but not "solved" (e.g., 2x+32x + 3).
  • Equations can be solved to find values of variables that make the statement true (e.g., 2x+3=72x + 3 = 7).
  • Key distinction: Presence of "==" defines an equation; absence means it is an expression.
  • Expressions have no truth value; equations are either true or false for given variable values.

Consequence: Recognizing this distinction clarifies whether the task is to simplify, evaluate, or solve.

Example: 4yβˆ’14y - 1 is an expression; 4yβˆ’1=114y - 1 = 11 is an equation.

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EXEC: ALGORITHM

Based on the core rules, which of the following mathematical statements is an equation?

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βœ–οΈ 5. Applications: Evaluating basic formulas in physics and geometry

πŸš€ Using Formulas

  • Velocity formula: v=dtv = \frac{d}{t} where dd is distance and tt is time.
  • Rectangle perimeter: P=2l+2wP = 2l + 2w where ll is length and ww is width.
  • Substitute known values into the formula to find the unknown.
  • Always include units in your final answer.

If distance is 100 meters and time is 5 seconds, then v=1005=20v = \frac{100}{5} = 20 meters per second.

πŸ’‘ Tip: Formulas are just expressions waiting for numbers!

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5. Applications: Evaluating basic formulas in physics and geometry

Applications: Evaluating Formulas in Physics and Geometry

Formulas are algebraic expressions or equations representing real-world relationships. Evaluation involves substituting measured or given values to compute unknown quantities.

Intuition: Formulas encode universal rules; substitution adapts them to specific scenarios.

Core Rules:

  • Velocity formula: v=dtv = \frac{d}{t} where vv is velocity, dd is distance, tt is time.
  • Perimeter of a rectangle: P=2l+2wP = 2l + 2w where ll is length and ww is width.
  • Substitute all known values and compute using order of operations.
  • Units matter: Ensure consistency (e.g., meters and seconds for velocity in meters per second).

Consequence: Mastery of substitution enables practical problem-solving across science and engineering.

Example: If d=150d = 150 meters and t=30t = 30 seconds, then v=15030=5v = \frac{150}{30} = 5 meters per second.

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A car travels a distance of d=300d = 300 meters in a time of t=15t = 15 seconds. Using the velocity formula, calculate the velocity of the car in meters per second.

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