Mathematics: Algebraic Reasoning – Grade 8

The Laws of Exponents serve as powerful tools for simplifying and manipulating algebraic expressions containing variables raised to various powers. Understanding these laws is essential for working efficiently with polynomial expressions and solving complex mathematical problems.

Before diving into the laws, it's crucial to remember that an exponent tells us how many times to multiply the base by itself. For example, $x^4$ means $x \cdot x \cdot x \cdot x$. This foundational understanding helps us derive and apply the exponent laws logically.

When working with variables, we apply the same principles we learned with numerical exponents. The difference is that we're working with unknown quantities, which makes the patterns even more elegant and powerful.

When multiplying expressions with the same base, we add the exponents. The Product Rule states: $x^a \cdot x^b = x^{a+b}$

For example:

• $x^5 \cdot x^8 = x^{5+8} = x^{13}$
• $y^{-3} \cdot y^7 = y^{-3+7} = y^4$
• $3z^4 \cdot 5z^2 = 15z^{4+2} = 15z^6$

This rule works because when we multiply $x^5 \cdot x^8$, we're actually multiplying $(x \cdot x \cdot x \cdot x \cdot x) \cdot (x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x)$, which gives us $x$ multiplied by itself $13$ times.

When dividing expressions with the same base, we subtract the exponents. The Quotient Rule states: $\frac{x^a}{x^b} = x^{a-b}$

For example:

• $\frac{x^9}{x^4} = x^{9-4} = x^5$
• $\frac{y^3}{y^8} = y^{3-8} = y^{-5} = \frac{1}{y^5}$
• $\frac{12m^7}{3m^2} = 4m^{7-2} = 4m^5$

This rule emerges from the fact that when we divide $\frac{x^9}{x^4}$, we can cancel out four $x$'s from both numerator and denominator, leaving us with five $x$'s in the numerator.

When raising a power to another power, we multiply the exponents. The Power Rule states: $(x^a)^b = x^{ab}$

For example:

• $(x^3)^4 = x^{3 \cdot 4} = x^{12}$
• $(y^{-2})^5 = y^{(-2) \cdot 5} = y^{-10}$
• $(2z^4)^3 = 2^3 \cdot (z^4)^3 = 8z^{12}$

This makes sense because $(x^3)^4$ means we're taking $x^3$ and using it as a factor four times: $x^3 \cdot x^3 \cdot x^3 \cdot x^3$, which by the Product Rule equals $x^{3+3+3+3} = x^{12}$.

Zero Exponent: Any non-zero base raised to the power of zero equals $1$: $x^0 = 1 \text{ (where } x \neq 0 \text{)}$

This follows from the pattern: $x^3 \div x^3 = \frac{x^3}{x^3} = 1$, and by the Quotient Rule, $x^3 \div x^3 = x^{3-3} = x^0$.

Negative Exponents: A negative exponent means we take the reciprocal: $x^{-n} = \frac{1}{x^n}$

For example:

• $x^{-4} = \frac{1}{x^4}$
• $\frac{1}{y^{-3}} = y^3$
• $3m^{-2} = \frac{3}{m^2}$

A monomial is an expression consisting of a number, a variable, or a product of numbers and variables with whole number exponents. When applying exponent laws to monomials, we work with both the numerical coefficients and the variable parts.

For example, to simplify $(3x^2y)(4x^3y^5)$:

1. Multiply coefficients: $3 \cdot 4 = 12$
2. Apply Product Rule to $x$: $x^2 \cdot x^3 = x^5$
3. Apply Product Rule to $y$: $y^1 \cdot y^5 = y^6$
4. Result: $12x^5y^6$

Mistake 1: Multiplying the base by the exponent instead of applying exponent rules.

• Incorrect: $x^3 \cdot x^2 = x^6$ (thinking $3 \times 2 = 6$)
• Correct: $x^3 \cdot x^2 = x^{3+2} = x^5$

Mistake 2: Adding exponents when multiplying different bases.

• Incorrect: $x^3 \cdot y^2 = xy^5$
• Correct: $x^3 \cdot y^2 = x^3y^2$ (cannot be simplified further)

Mistake 3: Confusion with negative exponents.

• Remember: $x^{-n}$ doesn't mean the result is negative; it means we take the reciprocal.

The Laws of Exponents appear frequently in scientific calculations, particularly when working with very large or very small numbers in scientific notation. For example, calculating areas and volumes often involves squaring and cubing expressions, while population growth and decay models use exponential expressions.

Understanding these laws also prepares you for more advanced topics like polynomial operations, rational expressions, and exponential functions that you'll encounter in later mathematics courses.

The distributive property is one of the most fundamental tools in algebra, enabling us to multiply expressions and transform them into equivalent forms. When working with linear expressions containing rational coefficients, this property becomes essential for simplifying complex mathematical relationships.

The distributive property states that multiplication distributes over addition and subtraction: $a(b + c) = ab + ac$

This property allows us to "distribute" a factor to each term inside parentheses. Think of it as sharing equally – if you have $a$ groups and each group contains $(b + c)$ items, the total is the same as having $ab$ items plus $ac$ items.

When multiplying a monomial (single term) by a polynomial (multiple terms), we distribute the monomial to each term in the polynomial.

Example 1: $3x(2x + 5)$

• Distribute $3x$ to each term: $3x \cdot 2x + 3x \cdot 5$
• Simplify: $6x^2 + 15x$

Example 2: $-4y(3y^2 - 2y + 7)$

• Distribute $-4y$: $(-4y)(3y^2) + (-4y)(-2y) + (-4y)(7)$
• Simplify: $-12y^3 + 8y^2 - 28y$

Rational coefficients include fractions and decimals. The distributive property works the same way, but we need to be careful with our arithmetic.

Example with Fractions: $\frac{2}{3}x(\frac{3}{4}x - \frac{1}{2})$

• Distribute: $\frac{2}{3}x \cdot \frac{3}{4}x + \frac{2}{3}x \cdot (-\frac{1}{2})$
• Multiply fractions: $\frac{2 \cdot 3}{3 \cdot 4}x^2 + \frac{2}{3} \cdot (-\frac{1}{2})x$
• Simplify: $\frac{6}{12}x^2 - \frac{2}{6}x = \frac{1}{2}x^2 - \frac{1}{3}x$

Example with Decimals: $0.25x(1.4x + 0.8)$

• Distribute: $0.25x \cdot 1.4x + 0.25x \cdot 0.8$
• Multiply: $0.35x^2 + 0.2x$

Area models provide a visual way to understand the distributive property. When we multiply $(x + 3)$ by $2$, we can think of it as finding the area of a rectangle.

|  x  |  3  |
|_____|_____|
| 2x  | 6   | 2
|_____|_____|

The total area is $2x + 6$, which matches our algebraic result: $2(x + 3) = 2x + 6$.

For more complex expressions like $x(2x + 5)$:

|  2x  |  5  |
|______|_____|
| 2x²  | 5x  | x
|______|_____|

Total area: $2x^2 + 5x$

When both expressions have multiple terms, we distribute each term in the first expression to each term in the second expression.

Example: $(2x + 3)(x + 4)$

• Distribute $2x$: $2x(x + 4) = 2x^2 + 8x$
• Distribute $3$: $3(x + 4) = 3x + 12$
• Combine: $2x^2 + 8x + 3x + 12 = 2x^2 + 11x + 12$

This can also be visualized with an area model:

|  x   |  4  |
|______|_____|
| 2x²  | 8x  | 2x
|______|_____|
| 3x   | 12  | 3
|______|_____|

Mistake 1: Forgetting to distribute to all terms.

• Incorrect: $3(x + 4) = 3x + 4$
• Correct: $3(x + 4) = 3x + 12$

Mistake 2: Errors with negative signs.

• When distributing a negative factor, remember that $-a(b + c) = -ab - ac$
• Example: $-2(x - 3) = -2x + 6$ (not $-2x - 6$)

Mistake 3: Combining unlike terms.

• Remember: $2x^2 + 3x$ cannot be simplified further because $x^2$ and $x$ are different types of terms.

After applying the distributive property, always look for like terms – terms with the same variable parts.

Example: $4x(2x + 3) + 5(x + 2)$

• Distribute: $8x^2 + 12x + 5x + 10$
• Combine like terms: $8x^2 + 17x + 10$

Like terms have the same variables raised to the same powers:

• $3x$ and $7x$ are like terms
• $2x^2$ and $-5x^2$ are like terms
• $3x$ and $3x^2$ are NOT like terms

The distributive property appears in many practical situations:

Shopping: If you buy $x$ shirts at 20$ each and $x$ pants at 35$each, your total cost is$20x + 35x = (20 + 35)x = 55x$.

Area Calculations: A rectangular garden has length $(2x + 5)$ feet and width $3$ feet. Its area is $3(2x + 5) = 6x + 15$ square feet.

Business Profit: If a company's daily revenue is $(100x + 50)$ dollars and daily costs are $(60x + 30)$ dollars, the profit is $(100x + 50) - (60x + 30) = 40x + 20$ dollars.

Mastering the distributive property with linear expressions prepares you for:

• Multiplying polynomials of higher degrees
• Factoring expressions (the reverse process)
• Solving quadratic equations
• Working with rational expressions
• Understanding function composition

The skills you develop here form the foundation for much of the algebra you'll encounter in advanced mathematics courses.

Factoring is the reverse of multiplication – instead of expanding expressions, we're breaking them down into their component parts. When expressions share common factors, we can rewrite them in a more compact and useful form that reveals important mathematical relationships.

Factoring means rewriting an expression as a product of its factors. It's the opposite of using the distributive property. While distribution takes us from $3(x + 4)$ to $3x + 12$, factoring takes us from $3x + 12$ back to $3(x + 4)$.

Factoring is like finding the "greatest common factor" but with algebraic expressions instead of just numbers. We're looking for the largest expression that divides evenly into each term.

Before factoring algebraic expressions, let's review finding the GCF of numbers:

• GCF of $12$ and $18$: Find factors of each
  • $12 = 2^2 \cdot 3$
  • $18 = 2 \cdot 3^2$
  • GCF = $2 \cdot 3 = 6$

For algebraic expressions, we find the GCF of both the numerical coefficients and the variable parts.

First, identify the GCF of all numerical coefficients in the expression.

Example: $15x^2 + 25x$

• Coefficients: $15$ and $25$
• $15 = 3 \cdot 5$ and $25 = 5^2$
• GCF of coefficients = $5$

For variables, the GCF is the variable raised to the smallest exponent that appears in all terms.

Example: $x^5 + x^3 + x^2$

• Variable parts: $x^5$, $x^3$, $x^2$
• Smallest exponent is $2$
• GCF of variable parts = $x^2$

Step 1: Identify the GCF of all terms Step 2: Factor out the GCF from each term Step 3: Write the result as GCF times the remaining expression Step 4: Verify by expanding back to the original

Example 1: Factor $12x^3 + 8x^2$

• Step 1: GCF of coefficients: $\gcd(12, 8) = 4$ GCF of variables: $\gcd(x^3, x^2) = x^2$ Overall GCF: $4x^2$
• Step 2: $12x^3 ÷ 4x^2 = 3x$ and $8x^2 ÷ 4x^2 = 2$
• Step 3: $12x^3 + 8x^2 = 4x^2(3x + 2)$
• Step 4: Check: $4x^2(3x + 2) = 12x^3 + 8x^2$ ✓

Example 2: Factor $18y^4 + 24y^2 - 6y$

• Step 1: GCF of coefficients: $\gcd(18, 24, 6) = 6$ GCF of variables: $\gcd(y^4, y^2, y) = y$ Overall GCF: $6y$
• Step 2: $18y^4 ÷ 6y = 3y^3$, $24y^2 ÷ 6y = 4y$, $6y ÷ 6y = 1$
• Step 3: $18y^4 + 24y^2 - 6y = 6y(3y^3 + 4y - 1)$

When expressions contain multiple variables, find the GCF for each variable separately.

Example: Factor $24x^3y^2 + 16x^2y^4 + 8xy$

• Coefficients: $\gcd(24, 16, 8) = 8$
• $x$ terms: $\gcd(x^3, x^2, x) = x$
• $y$ terms: $\gcd(y^2, y^4, y) = y$
• Overall GCF: $8xy$
• Result: $24x^3y^2 + 16x^2y^4 + 8xy = 8xy(3x^2y + 2xy^3 + 1)$

When working with fractions, find the GCF of the numerators and the LCM (least common multiple) of the denominators.

Example: Factor $\frac{3}{4}x^2 + \frac{1}{2}x$

• Convert to common denominators: $\frac{3}{4}x^2 + \frac{2}{4}x$
• GCF of numerators: $\gcd(3, 2) = 1$
• GCF including variable: $\frac{1}{4}x$
• Result: $\frac{3}{4}x^2 + \frac{1}{2}x = \frac{1}{4}x(3x + 2)$

Sometimes it's useful to factor out a negative common factor, especially when it makes the remaining expression simpler.

Example: Factor $-6x^2 + 4x$

• Option 1: $2x(-3x + 2)$
• Option 2: $-2x(3x - 2)$

Both are correct, but Option 2 might be preferred if it leads to a simpler form in the context of a larger problem.

When there's no common factor: If terms share no common factors other than 1, the expression is already in its simplest factored form.

Example: $3x^2 + 5y$ cannot be factored further because $3x^2$ and $5y$ share no common factors.

Perfect square terms: Be careful with expressions like $x^4 + x^2$:

• GCF is $x^2$
• Result: $x^2(x^2 + 1)$
• Note: $x^2 + 1$ cannot be factored further using real numbers

Factoring helps simplify algebraic fractions by revealing common factors in numerator and denominator.

Example: Simplify $\frac{15x^3 + 10x^2}{5x}$

• Factor numerator: $15x^3 + 10x^2 = 5x^2(3x + 2)$
• Simplify: $\frac{5x^2(3x + 2)}{5x} = \frac{x^2(3x + 2)}{x} = x(3x + 2) = 3x^2 + 2x$

Cost Analysis: A company's monthly costs are $50n + 75n$ where $n$ is the number of products. Factoring gives $n(50 + 75) = 125n$, showing the cost per product is $$$125$.

Geometry: The area of a rectangle is $6x^2 + 9x$. Factoring shows $3x(2x + 3)$, indicating possible dimensions of $3x$ by $(2x + 3)$.

Physics: In motion problems, expressions like $16t^2 + 24t$ factor to $8t(2t + 3)$, revealing time $t$ as a common factor in velocity calculations.

Mistake 1: Forgetting to factor out the complete GCF.

• Incorrect: $12x^3 + 8x^2 = 4(3x^3 + 2x^2)$
• Correct: $12x^3 + 8x^2 = 4x^2(3x + 2)$

Mistake 2: Errors in dividing terms by the GCF.

• When factoring $15x^3 + 10x^2$, remember $15x^3 ÷ 5x^2 = 3x$ (not $3x^2$)

Mistake 3: Not checking the answer by expanding.

• Always verify: If you factor $6x + 9$ as $3(2x + 3)$, check that $3(2x + 3) = 6x + 9$

Factoring by common factors is the first step toward more advanced factoring techniques:

• Factoring trinomials ($ax^2 + bx + c$)
• Factoring difference of squares ($a^2 - b^2$)
• Factoring perfect square trinomials
• Factoring by grouping

The skills you develop here – recognizing patterns, working systematically, and checking your work – will serve you well throughout your algebraic journey.

Multi-step linear equations represent some of the most practical and challenging problems in algebra. When variables appear on both sides of an equation, we must use systematic approaches to isolate the variable and find the solution.

A multi-step linear equation requires several operations to solve and may include:

• The distributive property
• Combining like terms
• Variables on both sides of the equation
• Rational coefficients (fractions and decimals)
• Parentheses and multiple grouping symbols

These equations model real-world situations where multiple factors influence the outcome, such as comparing different pricing plans, analyzing break-even points, or solving motion problems.

Step 1: Simplify both sides

• Use the distributive property to eliminate parentheses
• Combine like terms on each side

Step 2: Move all variable terms to one side

• Choose the side that will give you a positive coefficient
• Use addition or subtraction to move terms

Step 3: Move all constant terms to the opposite side

• Use addition or subtraction to isolate the variable term

Step 4: Divide by the coefficient of the variable

• This gives you the solution

Step 5: Check your solution

• Substitute back into the original equation

Example 1: $3(2x + 4) = 5x - 2$

Step 1: Simplify

• Left side: $3(2x + 4) = 6x + 12$
• Equation becomes: $6x + 12 = 5x - 2$

Step 2: Move variables to one side

• Subtract $5x$ from both sides: $6x - 5x + 12 = 5x - 5x - 2$
• Simplify: $x + 12 = -2$

Step 3: Move constants

• Subtract $12$ from both sides: $x + 12 - 12 = -2 - 12$
• Simplify: $x = -14$

Step 4: Check

• Left side: $3(2(-14) + 4) = 3(-28 + 4) = 3(-24) = -72$
• Right side: $5(-14) - 2 = -70 - 2 = -72$ ✓

Example 2: $\frac{1}{2}(4x - 6) + 3x = \frac{3}{4}x + 5$

Step 1: Simplify

• Left side: $\frac{1}{2}(4x) - \frac{1}{2}(6) + 3x = 2x - 3 + 3x = 5x - 3$
• Equation: $5x - 3 = \frac{3}{4}x + 5$

Step 2: Move variables (multiply by 4 to clear fractions)

• $4(5x - 3) = 4(\frac{3}{4}x + 5)$
• $20x - 12 = 3x + 20$
• $20x - 3x = 20 + 12$
• $17x = 32$
• $x = \frac{32}{17}$

Not all equations have exactly one solution. Understanding these special cases is crucial for complete algebraic fluency.

Infinitely Many Solutions When an equation simplifies to a true statement like $5 = 5$, every real number is a solution.

Example: $2(x + 3) = 2x + 6$

• Simplify: $2x + 6 = 2x + 6$
• Subtract $2x$: $6 = 6$ ✓
• This is always true, so the solution is all real numbers.

No Solution When an equation simplifies to a false statement like $3 = 7$, there is no solution.

Example: $3(x + 2) = 3x + 8$

• Simplify: $3x + 6 = 3x + 8$
• Subtract $3x$: $6 = 8$ ✗
• This is never true, so there is no solution.

Equations with fractions and decimals require extra care but follow the same principles.

Strategy 1: Clear fractions by multiplying by the LCD

Example: $\frac{1}{3}x + \frac{1}{6} = \frac{1}{2}x - \frac{1}{4}$

• LCD of denominators 3, 6, 2, 4 is 12
• Multiply everything by 12:
• $12 \cdot \frac{1}{3}x + 12 \cdot \frac{1}{6} = 12 \cdot \frac{1}{2}x - 12 \cdot \frac{1}{4}$
• $4x + 2 = 6x - 3$
• $4x - 6x = -3 - 2$
• $-2x = -5$
• $x = \frac{5}{2}$

Strategy 2: Work directly with decimals

Example: $0.3x + 1.2 = 0.5x - 0.8$

• $0.3x - 0.5x = -0.8 - 1.2$
• $-0.2x = -2$
• $x = \frac{-2}{-0.2} = 10$

Business Problem: Two cell phone plans cost the same total amount.

• Plan A: 25$ per month plus 0.05$ per text
• Plan B: 15$ per month plus 0.10$ per text
• When do they cost the same?

Let $t$ = number of texts

• Plan A cost: $25 + 0.05t$
• Plan B cost: $15 + 0.10t$
• Set equal: $25 + 0.05t = 15 + 0.10t$
• Solve: $25 - 15 = 0.10t - 0.05t$
• $10 = 0.05t$
• $t = 200$ texts

Both plans cost the same when you send 200 texts per month.

Mistake 1: Distribution errors

• Incorrect: $3(x + 4) = 3x + 4$
• Correct: $3(x + 4) = 3x + 12$

Mistake 2: Sign errors when moving terms

• When moving $-3x$ to the other side, add $3x$ to both sides
• When moving $+5$ to the other side, subtract $5$ from both sides

Mistake 3: Not checking the solution

• Always substitute your answer back into the original equation
• Both sides should give the same value

Mental Math Tips:

• When possible, choose to move variables to the side that will give a positive coefficient
• Look for opportunities to factor or use the distributive property to simplify
• Round decimal coefficients to check if your answer is reasonable

Graphing Calculator Check:

• Graph $y_1 = \text{left side}$ and $y_2 = \text{right side}$
• The x-coordinate of the intersection point is your solution
• If lines are parallel, there's no solution
• If lines coincide, there are infinitely many solutions

Mastering multi-step equations builds crucial problem-solving skills:

• Pattern Recognition: Identifying equation types quickly
• Strategic Thinking: Choosing the most efficient solution path
• Error Analysis: Finding and correcting mistakes systematically
• Verification: Developing habits of checking work

These skills transfer directly to scientific problem-solving, engineering calculations, and logical reasoning in many fields.

Linear inequalities extend our equation-solving skills to describe ranges of solutions rather than single values. Understanding inequalities is essential for modeling real-world constraints, optimization problems, and situations where we need to find all possible solutions within certain bounds.

Basic Inequality Symbols:

• $<$ means "less than" (strict inequality)
• $>$ means "greater than" (strict inequality)
• $\leq$ means "less than or equal to" (includes the boundary)
• $\geq$ means "greater than or equal to" (includes the boundary)

The key difference between equations and inequalities is that inequalities describe a range of solutions rather than a single value. For example, $x > 3$ means that $x$ can be any number greater than 3, such as 3.1, 4, 10, or 100.

Inequalities follow most of the same rules as equations, with one crucial exception:

Property 1: Addition and Subtraction Adding or subtracting the same value from both sides preserves the inequality:

• If $a < b$, then $a + c < b + c$ and $a - c < b - c$

Property 2: Multiplication and Division by Positive Numbers Multiplying or dividing both sides by a positive number preserves the inequality:

• If $a < b$ and $c > 0$, then $ac < bc$ and $\frac{a}{c} < \frac{b}{c}$

Property 3: Multiplication and Division by Negative Numbers ⚠️ CRITICAL: When multiplying or dividing both sides by a negative number, reverse the inequality symbol:

• If $a < b$ and $c < 0$, then $ac > bc$ and $\frac{a}{c} > \frac{b}{c}$

Why do we reverse the symbol? Consider the true statement $6 > -7$. If we multiply both sides by $-1$, we get $-6$ and $7$. Since $-6 < 7$, we must reverse the symbol to maintain truth.

The process for solving inequalities mirrors equation solving, with attention to the direction of the inequality symbol.

Example 1: $3x + 5 > 14$

• Subtract 5 from both sides: $3x > 9$
• Divide by 3 (positive): $x > 3$
• Solution: All numbers greater than 3

Example 2: $-2x + 7 \leq 15$

• Subtract 7 from both sides: $-2x \leq 8$
• Divide by -2 (negative, so reverse symbol): $x \geq -4$
• Solution: All numbers greater than or equal to -4

Example 3: $\frac{x}{4} - 3 < 2$

• Add 3 to both sides: $\frac{x}{4} < 5$
• Multiply by 4 (positive): $x < 20$
• Solution: All numbers less than 20

When variables appear on both sides, use the same systematic approach as with equations.

Example: $5x - 3 \geq 2x + 9$

• Subtract $2x$ from both sides: $3x - 3 \geq 9$
• Add 3 to both sides: $3x \geq 12$
• Divide by 3: $x \geq 4$

Visual representation helps us understand and verify inequality solutions.

Graphing Rules:

• Open circle (○): Use for $<$ and $>$ (the boundary point is NOT included)
• Closed circle (●): Use for $\leq$ and $\geq$ (the boundary point IS included)
• Shading direction: Shade toward the values that satisfy the inequality

Examples:

• $x > 3$: Open circle at 3, shade to the right →
• $x \leq -2$: Closed circle at -2, shade to the left ←
• $x \geq 0$: Closed circle at 0, shade to the right →

Example with Fractions: $\frac{2}{3}x - \frac{1}{4} > \frac{1}{2}$

Method 1: Clear fractions by multiplying by LCD (12)

• $12 \cdot \frac{2}{3}x - 12 \cdot \frac{1}{4} > 12 \cdot \frac{1}{2}$
• $8x - 3 > 6$
• $8x > 9$
• $x > \frac{9}{8}$

Method 2: Work directly with fractions

• $\frac{2}{3}x > \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$
• $x > \frac{3}{4} \div \frac{2}{3} = \frac{3}{4} \cdot \frac{3}{2} = \frac{9}{8}$

Budget Constraints: Maria has 150$ to spend on a party. Decorations cost 35, and each pizza costs $$$12. How many pizzas can she buy?

Let $p$ = number of pizzas

• Total cost: $35 + 12p$
• Constraint: $35 + 12p \leq 150$
• Solve: $12p \leq 115$
• $p \leq 9.58...$
• Since she can't buy part of a pizza: $p \leq 9$

Maria can buy at most 9 pizzas.

Sales Goals: A salesperson earns 300$ per week plus 25 per sale. How many sales are needed to earn at least $$$750 per week?

Let $s$ = number of sales

• Weekly earnings: $300 + 25s$
• Goal: $300 + 25s \geq 750$
• Solve: $25s \geq 450$
• $s \geq 18$

The salesperson needs at least 18 sales per week.

Sometimes we need to express that a variable is between two values:

• $-3 < x < 7$ means $x$ is greater than -3 AND less than 7
• This is equivalent to: $x > -3$ AND $x < 7$

Interval notation provides a compact way to write solution sets:

• $(a, b)$: All numbers between $a$ and $b$ (not including endpoints)
• $[a, b]$: All numbers between $a$ and $b$ (including endpoints)
• $(a, \infty)$: All numbers greater than $a$
• $(-\infty, b]$: All numbers less than or equal to $b$

Examples:

• $x > 3$ can be written as $(3, \infty)$
• $x \leq -2$ can be written as $(-\infty, -2]$
• $-1 \leq x < 4$ can be written as $[-1, 4)$

Mistake 1: Forgetting to reverse the inequality symbol

• Always remember: When multiplying or dividing by a negative number, flip the symbol
• Double-check by testing a value in your solution

Mistake 2: Confusing open and closed circles

• $<$ and $>$ use open circles (boundary not included)
• $\leq$ and $\geq$ use closed circles (boundary included)

Mistake 3: Shading the wrong direction

• After plotting the boundary point, test a value to see which side satisfies the inequality
• For $x > 3$, test $x = 4$: Since $4 > 3$ is true, shade to the right of 3

Method 1: Test boundary values

• For $x \geq -2$, test $x = -2$: Should satisfy the original inequality
• Test a value just outside: $x = -3$ should NOT satisfy the inequality

Method 2: Test a value in your solution set

• For $x < 5$, test $x = 0$: Substitute into the original inequality to verify

Method 3: Graph both sides

• Graph the left and right sides of the inequality as separate functions
• The solution is where one function is above/below the other

Mastering linear inequalities prepares you for:

• Compound inequalities involving AND/OR logic
• Absolute value inequalities with multiple solution regions
• Systems of inequalities with overlapping solution regions
• Linear programming in optimization problems
• Quadratic inequalities in advanced algebra

The logical reasoning skills you develop with inequalities are fundamental to mathematical modeling and real-world problem solving.

Equations involving squares and cubes represent a special class of algebraic problems that require understanding of inverse operations and the properties of roots. These equations appear frequently in geometry, physics, and real-world applications involving areas, volumes, and quadratic relationships.

A square root equation has the form $x^2 = p$, where $p$ is a whole number. To solve these equations, we use the square root operation, which is the inverse of squaring.

Key Concept: When we take the square root of both sides of $x^2 = p$, we must consider both the positive and negative square roots because both $(\sqrt{p})^2 = p$ and $(-\sqrt{p})^2 = p$.

The complete solution is: $x = \pm\sqrt{p}$

This is read as "x equals plus or minus the square root of p."

Example 1: $x^2 = 25$

• Take the square root of both sides: $x = \pm\sqrt{25}$
• Simplify: $x = \pm 5$
• Solutions: $x = 5$ or $x = -5$
• Check: $5^2 = 25$ ✓ and $(-5)^2 = 25$ ✓

Example 2: $x^2 = 49$

• $x = \pm\sqrt{49} = \pm 7$
• Solutions: $x = 7$ or $x = -7$

Example 3: $x^2 = 100$

• $x = \pm\sqrt{100} = \pm 10$
• Solutions: $x = 10$ or $x = -10$

For Grade 8, you should know perfect squares up to $15^2 = 225$:

A cube root equation has the form $x^3 = q$, where $q$ is an integer (positive, negative, or zero). Unlike square roots, cube roots can be negative, and each cube has exactly one real cube root.

The solution is: $x = \sqrt[3]{q}$

Important difference: Cube root equations have exactly one real solution, while square root equations typically have two solutions.

Example 1: $x^3 = 27$

• Take the cube root: $x = \sqrt[3]{27}$
• Simplify: $x = 3$
• Check: $3^3 = 27$ ✓

Example 2: $x^3 = -8$

• Take the cube root: $x = \sqrt[3]{-8}$
• Simplify: $x = -2$
• Check: $(-2)^3 = -8$ ✓

Example 3: $x^3 = 0$

• $x = \sqrt[3]{0} = 0$
• Check: $0^3 = 0$ ✓

For Grade 8, know perfect cubes from $(-5)^3 = -125$ to $5^3 = 125$:

Visual Understanding:

• Squaring relates to area: A square with side length $s$ has area $s^2$
• Cubing relates to volume: A cube with side length $s$ has volume $s^3$

Numerical Patterns:

• Doubling a number vs. squaring it:
  • $2 \times 3 = 6$ (doubling)
  • $3^2 = 9$ (squaring)
• Tripling a number vs. cubing it:
  • $3 \times 4 = 12$ (tripling)
  • $4^3 = 64$ (cubing)

Geometry - Square Areas: A square garden has an area of 144 square feet. What is the length of each side?

• Let $s$ = side length
• Area equation: $s^2 = 144$
• Solve: $s = \pm\sqrt{144} = \pm 12$
• Since length must be positive: $s = 12$ feet

Geometry - Cube Volumes: A cube-shaped storage container has a volume of 216 cubic inches. What is the length of each edge?

• Let $e$ = edge length
• Volume equation: $e^3 = 216$
• Solve: $e = \sqrt[3]{216}$
• Since $6^3 = 216$: $e = 6$ inches

Physics - Free Fall: The distance $d$ (in feet) an object falls in $t$ seconds is given by $d = 16t^2$. How long does it take for an object to fall 400 feet?

• Equation: $16t^2 = 400$
• Divide by 16: $t^2 = 25$
• Solve: $t = \pm 5$
• Since time must be positive: $t = 5$ seconds

When perfect squares or cubes aren't immediately obvious, use systematic approaches:

For square roots:

• Estimate by finding perfect squares nearby
• Example: $\sqrt{150}$ is between $\sqrt{144} = 12$ and $\sqrt{169} = 13$

For cube roots:

• Look for patterns in the ones digit
• Numbers ending in 7 have cubes ending in 3: $7^3 = 343$
• Numbers ending in 3 have cubes ending in 7: $3^3 = 27$

When there are no real solutions:

• $x^2 = -9$ has no real solutions because squares of real numbers are never negative
• However, $x^3 = -27$ does have a real solution: $x = -3$

Working with irrational solutions:

• $x^2 = 50$ gives $x = \pm\sqrt{50} = \pm 5\sqrt{2}$
• These are exact answers; decimal approximations like $\pm 7.07$ are less precise

Always verify by substitution:

For $x^2 = 81$ with solution $x = \pm 9$:

• Check $x = 9$: $9^2 = 81$ ✓
• Check $x = -9$: $(-9)^2 = 81$ ✓

For $x^3 = -64$ with solution $x = -4$:

• Check: $(-4)^3 = -64$ ✓

Mistake 1: Forgetting the negative solution for square roots

• Incorrect: $x^2 = 36$ so $x = 6$
• Correct: $x^2 = 36$ so $x = \pm 6$

Mistake 2: Thinking cube roots are always positive

• Remember: $\sqrt[3]{-8} = -2$, not $2$

Mistake 3: Confusing squaring/doubling and cubing/tripling

• $3^2 = 9$ (not $6$)
• $2^3 = 8$ (not $6$)

Calculator skills:

• Most calculators have $\sqrt{x}$ and $\sqrt[3]{x}$ functions
• Remember that calculators typically give only the positive square root
• For $x^2 = 25$, calculator gives $\sqrt{25} = 5$, but you must remember $x = \pm 5$

Estimation skills:

• Develop number sense for common perfect squares and cubes
• Use mental math to check if calculator results are reasonable

Mastering square and cube root equations prepares you for:

• Quadratic equations using the quadratic formula
• Radical equations with square roots in the equation itself
• Exponential and logarithmic equations as inverse operations
• Pythagorean theorem applications in geometry
• Complex numbers for equations like $x^2 = -1$

The inverse operation thinking you develop here is fundamental to understanding mathematical relationships and solving increasingly complex problems.

Understanding the relationship between linear and proportional relationships is crucial for mathematical modeling and problem-solving. While all proportional relationships are linear, not all linear relationships are proportional – and knowing the difference helps you choose appropriate mathematical tools and make accurate predictions.

Linear Relationship: A relationship where the rate of change between two variables is constant. When graphed, it forms a straight line.

Proportional Relationship: A special type of linear relationship where the ratio between corresponding values of two variables is constant, and the relationship passes through the origin (0, 0).

Key Insight: All proportional relationships are linear, but not all linear relationships are proportional.

The fundamental distinction lies in whether the line passes through the origin:

Proportional Relationship:

• Always passes through (0, 0)
• Has equation form $y = mx$ (no y-intercept term)
• Constant ratio: $\frac{y}{x} = m$ for all points

Non-Proportional Linear Relationship:

• Does NOT pass through (0, 0)
• Has equation form $y = mx + b$ where $b \neq 0$
• Constant rate of change but not constant ratio

Method 1: Check if (0, 0) is included or implied

Example 1 - Proportional:

• Passes through (0, 0) ✓
• Constant ratio: $\frac{30}{2} = \frac{60}{4} = \frac{90}{6} = 15$ ✓
• This IS proportional: Pay = $15 × Hours

Example 2 - Linear but Not Proportional:

• Does NOT pass through (0, 0) – there's a $50 base pay ✗
• Constant rate of change: $15 per hour ✓
• This is linear but NOT proportional: Pay = $15 × Hours +$50

Method 2: Calculate ratios and check for consistency

For a relationship to be proportional, $\frac{y}{x}$ must be the same for all data points (excluding where $x = 0$).

Visual Inspection:

• Proportional: Line passes through origin (0, 0)
• Non-Proportional Linear: Line is straight but does not pass through origin

What if the origin isn't shown? Use two points to find the equation, then check if substituting $x = 0$ gives $y = 0$.

Example: Points (2, 8) and (5, 14)

• Slope: $m = \frac{14 - 8}{5 - 2} = \frac{6}{3} = 2$
• Using point-slope form: $y - 8 = 2(x - 2)$
• Simplify: $y = 2x + 4$
• When $x = 0$: $y = 4 \neq 0$
• Therefore, NOT proportional

Proportional equations have the form $y = mx$:

• $y = 3x$ ✓ Proportional
• $C = \pi d$ ✓ Proportional (circumference and diameter)
• $d = 65t$ ✓ Proportional (distance and time at constant speed)

Non-proportional linear equations have the form $y = mx + b$ where $b \neq 0$:

• $y = 2x + 5$ ✗ Not proportional (y-intercept is 5)
• $F = \frac{9}{5}C + 32$ ✗ Not proportional (Fahrenheit-Celsius conversion)
• $y = -3x - 7$ ✗ Not proportional (y-intercept is -7)

Proportional Situations:

• Unit pricing: Total cost vs. quantity (when there's no fixed fee)
• Distance vs. time: At constant speed with no initial displacement
• Recipe scaling: Ingredient amounts vs. number of servings

Example: "Bananas cost $1.50 per pound"

• Cost = $1.50 × Pounds
• No initial cost, so it passes through (0, 0)
• This IS proportional

Non-Proportional Linear Situations:

• Cell phone plans: Monthly cost with base fee plus per-minute charges
• Taxi fares: Base fare plus cost per mile
• Temperature conversions: Celsius to Fahrenheit

Example: "Cell phone plan costs $25 per month plus$0.10 per text"

• Cost = $25 +$0.10 × Texts
• Even with 0 texts, cost is $25 (doesn't pass through origin)
• This is linear but NOT proportional

These three concepts are related but have important distinctions:

Unit Rate: The amount of change in the dependent variable per one unit of the independent variable.

Constant of Proportionality: In a proportional relationship $y = mx$, the value $m$ is the constant of proportionality.

Slope: The rate of change in any linear relationship, proportional or not.

Key Relationship:

• In proportional relationships: Unit Rate = Constant of Proportionality = Slope
• In non-proportional linear relationships: Unit Rate = Slope ≠ Constant of Proportionality (which doesn't exist)

Misconception 1: "All linear relationships are proportional" Truth: Only linear relationships that pass through the origin are proportional.

Misconception 2: "If there's a constant rate of change, it's proportional" Truth: Constant rate of change indicates linearity, but proportionality requires the additional condition of passing through the origin.

Misconception 3: "The origin must be visible on the graph to determine proportionality" Truth: You can determine proportionality by finding the equation and checking if the y-intercept is zero.

Strategy 1: The Zero Test Ask: "When the independent variable is zero, is the dependent variable also zero?"

• If yes → Could be proportional (check for constant ratio)
• If no → Definitely not proportional (but could still be linear)

Strategy 2: The Ratio Test Calculate $\frac{y}{x}$ for several data points:

• If all ratios are equal → Proportional
• If ratios are not equal → Not proportional

Strategy 3: The Equation Test Find the linear equation:

• Form $y = mx$ → Proportional
• Form $y = mx + b$ where $b \neq 0$ → Not proportional

Science: Hooke's Law ($F = kx$) is proportional – the force needed to stretch a spring is proportional to the displacement.

Economics: Simple interest ($I = Prt$) shows a proportional relationship between interest and time (when principal and rate are constant).

Geometry: The relationship between circumference and diameter ($C = \pi d$) is proportional with constant of proportionality $\pi$.

Sports: A runner maintaining constant speed creates a proportional relationship between distance and time.

Understanding the distinction between proportional and linear relationships develops several important mathematical thinking skills:

Pattern Recognition: Identifying whether relationships maintain constant ratios or just constant rates

Critical Analysis: Not accepting that correlation implies proportionality

Modeling Decisions: Choosing appropriate mathematical models based on the nature of relationships

Contextual Interpretation: Understanding when real-world situations involve fixed costs, base amounts, or offset values

These analytical skills transfer to more advanced mathematics where distinguishing between different types of relationships becomes increasingly important for accurate modeling and problem-solving.