Understanding Linear Equations

• Linear equations are fundamental in mathematics, representing relationships between variables with a constant rate of change
• They form the basis for more complex algebraic concepts and are widely used in various fields

Definition of a Linear Equation

• A linear equation is an algebraic equation where each term is either a constant or a product of a constant and a single variable.
• It is generally written as ax + b = 0, where x is the variable, a and b are constants, and "a" is not equal to zero.

Types of Linear Equations

• Linear equations can involve one or more variables.
• Equations with one variable can be solved directly.
• Equations with two or more variables represent lines, planes, or hyperplanes when graphed.

Solving Linear Equations with One Variable

• To solve a linear equation with one variable, isolate the variable on one side using algebraic operations.
• Operations include addition, subtraction, multiplication, and division, applied equally to both sides.
• Example: Solve 3x + 5 = 14
  • Subtract 5 from both sides: 3x = 9
  • Divide by 3: x = 3

Solving Linear Equations with Multiple Variables

• Equations with multiple variables, such as 2x + 3y = 6, show a relationship between the variables.
• They have infinite solutions, representable graphically as a straight line.
• Equations can be rearranged to express one variable in terms of another, for example, y = (-2/3)x + 2.

Systems of Linear Equations

• A system of linear equations includes two or more linear equations with the same variables.
• The solution is the set of values that satisfy all equations simultaneously.
• Systems can have one, none, or infinitely many solutions.
• Methods for solving include substitution, elimination, and matrix methods.

Solving Systems of Linear Equations by Substitution

• Solve one equation for one variable and substitute into the other equation.
• This gives a single equation with one variable, which can be solved.
• Substitute the value back into an original equation to find the other variable's value.
• Example:
  • Solve the system:
    • x + y = 5
    • 2x - y = 1
  • Express y in terms of x: y = 5 - x
  • Substitute into the second equation: 2x - (5 - x) = 1
  • Simplify and solve for x: 3x - 5 = 1 => 3x = 6 => x = 2
  • Substitute x = 2 back into y = 5 - x: y = 5 - 2 = 3
  • The solution is x = 2, y = 3

Solving Systems of Linear Equations by Elimination

• Manipulate equations so coefficients of one variable are the same or additive inverses.
• Adding or subtracting the equations eliminates one variable.
• Solve the resulting equation, and substitute back to find the other variable.
• Example:
  • Solve the system:
    • 3x + 2y = 7
    • 4x - 2y = 0
  • The coefficients of y are additive inverses.
  • Add the equations to eliminate y: 7x = 7 => x = 1
  • Substitute x = 1 back into 3x + 2y = 7
  • 3(1) + 2y = 7 => 2y = 4 => y = 2
  • The solution is x = 1, y = 2

Applications of Linear Equations

• They model relationships between quantities like distance/time, cost/quantity, or supply/demand.
• Used to solve problems involving mixtures, rates, and proportions.
• They assist in analyzing data, making predictions and optimizing processes in various fields.

Graphing Linear Equations

• Linear equations in two variables are graphed on a coordinate plane.
• The graph is always a straight line.
• Graph by finding and plotting at least two points that satisfy the equation.
• The x-intercept is where the line crosses the x-axis (y = 0).
• The y-intercept is where the line crosses the y-axis (x = 0).
• The slope represents the rate of change of y with respect to x.

Slope-Intercept Form

• The slope-intercept form is y = mx + b, where:
  • m is the slope (rate of change of y with respect to x).
  • b is the y-intercept (where the line crosses the y-axis, x = 0).
• This form helps identify the slope and y-intercept for graphing or analysis.

Point-Slope Form

• The point-slope form is y - y1 = m(x - x1), where:
  • (x1, y1) is a known point on the line.
  • m is the slope of the line.
• Useful when a point and slope are known and the equation is needed.

Parallel and Perpendicular Lines

• Parallel lines have the same slope (m1 = m2).
• Perpendicular lines have slopes that are negative reciprocals.
• For perpendicular lines, m1 * m2 = -1, or m2 = -1/m1.

Linear Inequalities

• A linear inequality uses an inequality sign (<, >, ≤, ≥) instead of an equals sign.
• Solving involves finding the set of values that satisfy the inequality.
• The solution is expressed as an interval on a number line or a region on a coordinate plane.
• Multiplying or dividing by a negative number reverses the inequality sign.

Graphing Linear Inequalities

• Graph the corresponding linear equation (replace inequality with equals sign).
  • Use a dashed line for strict inequalities (< or >).
  • Use a solid line for non-strict inequalities (≤ or ≥).
• Choose a test point not on the line and substitute into the original inequality.
• Shade the region on the same side as the test point if it satisfies the inequality.
• Shade the opposite side if the test point does not satisfy the inequality.

Absolute Value Equations

• Absolute value equations contain an absolute value expression.
• The absolute value is the distance from zero and is always non-negative.
• Solve by considering two cases:
  • The expression inside the absolute value is positive or zero.
  • The expression inside the absolute value is negative.
• Example: Solve |x - 3| = 5:
  • Case 1: x - 3 = 5 => x = 8
  • Case 2: x - 3 = -5 => x = -2
  • The solutions are x = 8 and x = -2

Absolute Value Inequalities

• Absolute value inequalities contain an absolute value expression.
• Solve by considering two cases, similar to absolute value equations.
• Example: Solve |x - 2| < 3:
  • Case 1: x - 2 < 3 => x < 5
  • Case 2: -(x - 2) < 3 => -x + 2 < 3 => -x < 1 => x > -1
  • The solution is -1 < x < 5, or the interval (-1, 5)
• Example: Solve |2x + 1| ≥ 5:
  • Case 1: 2x + 1 ≥ 5 => 2x ≥ 4 => x ≥ 2
  • Case 2: -(2x + 1) ≥ 5 => -2x - 1 ≥ 5 => -2x ≥ 6 => x ≤ -3
  • The solution is x ≤ -3 or x ≥ 2

Expanding Brackets

• Expanding brackets is a fundamental algebraic skill used to simplify expressions.
• It involves multiplying each term inside the bracket by the term outside the bracket.

Basic Expansion: Single Term Outside

• a(b + c) = ab + ac: This constitutes the distributive property.
• The term outside the bracket (a) is multiplied by each term inside the bracket (b and c).
• Example: 3(x + 2) = 3x + 6, the 3 is multiplied by both x and 2.
• Example with subtraction: 4(y - 5) = 4y - 20, the 4 is multiplied by y and -5.
• Brackets can contain more than two terms. Example: a(b + c + d) = ab + ac + ad
• Example: 2(x + 3 + z) = 2x + 6 + 2z, the 2 is multiplied by x, 3, and z.
• Expanding with negative terms outside the bracket requires special attention to signs.
• A negative outside the bracket changes the sign of every term inside the bracket.
• Example: -2(x + 3) = -2x - 6, note how +3 becomes -6 after multiplication by -2.
• Example: -a(b - c) = -ab + ac, The -a multiplies both b and -c, changing their signs accordingly.

Expanding Two Binomials: (a + b)(c + d)

• Expanding two binomials involves multiplying each term in the first bracket by each term in the second bracket.
• (a + b)(c + d) = ac + ad + bc + bd; this is often remembered using the FOIL method.
• FOIL stands for First, Outer, Inner, Last, referring to the order of multiplications.
  • First: Multiply the first terms in each bracket (a and c).
  • Outer: Multiply the outer terms (a and d).
  • Inner: Multiply the inner terms (b and c).
  • Last: Multiply the last terms in each bracket (b and d).
• Example: (x + 2)(x + 3) = xx + x3 + 2x + 23 = x² + 3x + 2x + 6
• Simplify the expanded expression by combining like terms: x² + 5x + 6.
• Example with negative terms: (x - 4)(x + 1) = xx + x1 + (-4)*x + (-4)*1 = x² + x - 4x - 4
• Simplify: x² - 3x - 4.

Expanding Special Cases

• Squaring a binomial: (a + b)² = (a + b)(a + b) = a² + 2ab + b²
• (a - b)² = (a - b)(a - b) = a² - 2ab + b²
• Example: (x + 3)² = (x + 3)(x + 3) = x² + 3x + 3x + 9 = x² + 6x + 9
• Example: (y - 2)² = (y - 2)(y - 2) = y² - 2y - 2y + 4 = y² - 4y + 4
• Difference of two squares: (a + b)(a - b) = a² - b²
• Example: (x + 4)(x - 4) = x² - 4², which simplifies to x² - 16.

Expanding More Complex Expressions

• Expanding expressions with multiple brackets requires careful application of the distributive property in stages.
• Example: 2(x + 1)(x + 3). First, expand the two binomials: (x + 1)(x + 3) = x² + 4x + 3
• Then, multiply the result by 2: 2(x² + 4x + 3) = 2x² + 8x + 6
• Expressions can involve a combination of single-term brackets and binomial brackets.
• Example: x(x + 2) + 3(x - 1). First, expand each bracket separately.
• x(x + 2) = x² + 2x, and 3(x - 1) = 3x - 3; then, combine the results: x² + 2x + 3x - 3, which simplifies to x² + 5x - 3.

Common Mistakes

• Forgetting to multiply every term inside the bracket: e.g., incorrectly expanding a(b + c) as ab instead of ab + ac.
• Errors with signs, especially when dealing with negative terms outside or inside the brackets.
• Incorrectly squaring a binomial, thinking (a + b)² = a² + b² instead of a² + 2ab + b², where the middle term (2ab) is often missed.
• Not combining like terms after expanding.

Applications

• Expanding brackets is crucial for solving equations.
• It helps simplify expressions before solving for unknown variables.
• It is used in rearranging formulas.
• Expanding brackets is used in simplifying algebraic fractions.
• It is essential in calculus, particularly when differentiating or integrating certain functions.
• Factoring is the reverse process of expanding, and both are vital tools.