Comprehensive Study Notes on Linear Equations, Formulas, and Applications

Linear Equations in One Variable

• An equation is a mathematical statement containing an equal sign ($=$), setting two expressions equal to each other.
• Solutions (also referred to as roots) are values that satisfy the equation, making it a true statement.
• A solution set is the collection of all numbers that satisfy a given equation.

Fundamental Properties of Equality

• Addition Property of Equality: If $a = b$, then $a + c = b + c$. Adding the exact same quantity to both sides of an equation maintains equality.
• Multiplication Property of Equality: If $a = b$ and $c \neq 0$, then $a \times c = b \times c$. Multiplying both sides of an equation by the same non-zero quantity maintains equality.

Linear vs. Non-Linear Equations

• A linear equation in one variable (also called a first-degree equation) is an equation that can be written in the standard form:   $a \times x + b = 0$   where $a$ and $b$ are real numbers and a \neq 0$.\n* It is classified as a first-degree equation because the variable x$is raised to the power of$1.\n* Non-linear equations contain operations or exponents that violate the first-degree structure. Examples of non-linear equations include:\n * Equations containing square roots of variables: \sqrt{x + 2} = 5\n * Equations dividing by a variable: \frac{1}{x}\n * Equations with squared variables: x^2 + 3 \times x\n\n## Solving Linear Equations\n\n### Example 1: Applying Distributive Property and Sign Changes\n\nTo solve linear equations involving parentheses:\n1. Apply the distributive property to eliminate parentheses.\n2. Pay strict attention to negative signs. A negative sign preceding parentheses acts as multiplying by -1, changing the sign of every term inside.\n\nFor the equation solved step-by-step:\n* Distribute coefficients and sign changes across parentheses:\n  6 \times x - 12 = 7 - x - 5\n* Combine like terms on the right side (7 - 5 = 2):\n  6 \times x - 12 = 2 - x\n* Add x to both sides using the Addition Property of Equality:\n  7 \times x - 12 = 2\n* Add 12 to both sides:\n  7 \times x = 14\n* Divide both sides by 7:\n  x = 2\n\n#### Checking Solution x = 2\n* Substitute x = 2 back into the original expressions:\n  3 \times (2 \times 2 - 3) = 7 - (2 + 5)\n  3 \times (4 - 3) = 7 - 7\n  0 = 0\n* Since 0 = 0 is a true statement, the solution checks out.\n* The solution set is {2}.\n\n### Example 2: Equations Containing Fractions\n\nWhen solving linear equations with fractional terms:\n1. Find the least common denominator (LCD) for all denominators in the equation.\n2. Multiply every term on both sides of the equation by the LCD to clear all fractions.\n\nFor an equation with denominators 3$,$2$, and$4$, the common denominator is$12.\n\n* Multiply the entire equation by 12:\n  12 \times \left(\frac{2 \times x + 3}{3}\right) + 12 \times \left(\frac{1}{2} \times x\right) = 12 \times \left(\frac{1}{4} \times x\right) - 12 \times \left(\frac{7}{3}\right)\n* Simplify each term after multiplication:\n  4 \times (2 \times x + 4) + 6 \times x = 3 \times x - 28\n  8 \times x + 16 + 6 \times x = 3 \times x - 28\n* Combine like terms:\n  14 \times x + 16 = 3 \times x - 28\n* Subtract 3 \times x from both sides:\n  11 \times x + 16 = -28\n* Subtract 16 from both sides:\n  11 \times x = -44\n* Divide both sides by 11:\n  x = -4\n\n#### Checking Solution x = -4\n* Substitute x = -4 back into the original equation:\n  \frac{2 \times (-4) + 4}{3} + \frac{1}{2} \times (-4) = \frac{1}{4} \times (-4) - \frac{7}{3}\n  \frac{-4}{3} + (-2) = -1 - \frac{7}{3}\n* Obtain a common denominator of 3 on both sides:\n  -\frac{4}{3} - \frac{6}{3} = -\frac{3}{3} - \frac{7}{3}\n  -\frac{10}{3} = -\frac{10}{3}\n* The equality holds true, verifying that x = -4 is correct.\n\n# Classifications of Linear Equations\n\nEquations are categorized into three distinct types based on the nature of their solution sets:\n\n## 1. Identity\n* An **identity** is an equation that is true for all possible real values of the variable.\n* Solving an identity algebraically results in a universally true statement such as 0 = 0$.
• Solution Set: All real numbers.

2. Conditional Equation

• A conditional equation is an equation that is true under specific conditions (i.e., for a limited number of specific values).
• Example 1:   $2 \times x = 4 \implies x = 2$
• Example 2:   $5 \times x - 4 = 11$$5 \times x = 15$$x = 3$
• Solution Set: A finite set containing the specific root, such as $\{3\}$.

3. Contradiction

• A contradiction is an equation that is false for all values of the variable; it has no solution.
• Example 1:   $x = x + 1$   If $x = 3$, then $3 = 4$, which is false.
• Example 2:   $3 \times (3 \times x - 1) = 9 \times x + 7$$9 \times x - 3 = 9 \times x + 7$   Subtract $9 \times x$ from both sides:   $-3 = 7$   Since $-3 = 7$ is a false statement, the equation is a contradiction.
• Solution Set: The empty set or null set, denoted by the symbol $\emptyset$ (a circle with a slash line through it).

Summary Table of Equation Types

Formulas and Literal Equations

Literal equations are formulas containing multiple variables. Solving a literal equation for a specific variable involves isolating that variable on one side of the equal sign.

Simple Interest Formula

The simple interest formula is expressed as: $I = P \times r \times t$ where:

• $I$ = Interest earned or paid
• $P$ = Principal amount
• $r$ = Annual simple interest rate (expressed as a decimal)
• $t$ = Time (in years)

Solving Simple Interest for Time ($t$)

To isolate $t$ when $I = P \times r \times t$:

1. Divide both sides by $P \times r$:    $t = \frac{I}{P \times r}$

Solving for Principal ($P$) in Accumulated Amount Formulas

Given the total accumulated amount formula $A = P + P \times r \times t$:

1. Identify $P$ as a common factor on the right side.
2. Factor out $P$ using the distributive property:    $A = P \times (1 + r \times t)$
3. Divide both sides by $(1 + r \times t)$:    $P = \frac{A}{1 + r \times t}$

Solving Multi-Variable Expressions for $x$

When working with expressions containing multiple constants ($a$, $b$):

1. Apply the distributive property to remove parentheses.
2. Group all terms containing $x$ on one side of the equation and move constant terms ($a$, $b$) to the opposite side.
3. Combine like terms. For example, subtracting $4 \times x$ from $6 \times x$ leaves 2 \times x$.\n4. Add 15 \times a$to both sides and subtract$4 \times b from both sides:\n   2 \times x = 15 \times a - 4 \times b\n5. Divide all terms by 2:\n   x = \frac{15 \times a - 4 \times b}{2}\n\n# Real-World Applications: Simple Interest Calculation\n\n## Problem Statement\nA borrower takes out a loan of \$5,240$for new furniture and plans to pay it off in$11$months at an annual simple interest rate of$4.5\%. Calculate the total interest paid.\n\n## Step-by-Step Solution\n\n1. **Identify Given Values**:\n * Principal (P$):$\$5,240\n * Rate (r$):$4.5\%\n * Time (t$):$11 months\n\n2. **Convert Units**:\n * Convert interest rate r to a decimal by moving the decimal point two places to the left:\n     r = 0.045\n * Convert time t$from months to years using the conversion factor$12\,\text{months} = 1\,\text{year}. Set time in years as:\n     t = \frac{11}{12}\n\n3. **Substitute into Formula**:\n   I = P \times r \times t\n   I = 5240 \times 0.045 \times \left(\frac{11}{12}\right)\n\n4. **Perform Calculation**:\n   I = 235.8 \times \left(\frac{11}{12}\right)\n   I = 216.15\n\n5. **Final Result**:\n   The total interest paid on the furniture loan is \$216.15$$

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