ALG-1: Order of Operations and Algebraic Structure
Module 2 · Algebraic Manipulation
Section 1: Introduction
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Quick question. No calculator, no tricks — just tell me: what does \( -3^2 \) equal?
If you said 9, you're in good company. Most students do. But the correct answer is \( -9 \).
That's not a trick. It's not a typo. It's how algebra is read — and if you misread it, your calculus work will silently go wrong every single time this pattern appears. And it appears constantly.
Here's why: \( -3^2 \) means the negative of \( 3^2 \). The exponent grabs only the \( 3 \), not the negative sign. So \( -3^2 = -(3^2) = -(9) = -9 \). The expression \( (-3)^2 \) is a different thing: now the negative is inside the base, giving \( (-3)(-3) = 9 \).
This isn't an isolated quirk. It's a symptom of something deeper: algebraic notation is a structured language with precise grammar rules, and when students skip or misremember those rules, errors compound. The fraction bar hides invisible parentheses. The minus sign has reach — but only sometimes. Exponents bind tightly. Operations don't simply flow left to right.
This lesson builds the habit of reading algebraic structure correctly. Once internalized, these rules will feel automatic — and a whole class of calculus errors simply won't happen to you.
After this lesson, you will be able to:
Apply BEDMAS/PEMDAS correctly, including left-to-right resolution of same-precedence operations
Recognize the fraction bar as a grouping symbol that brackets both numerator and denominator
Work from the innermost grouping outward in nested expressions
Distinguish \( -x^2 \) from \( (-x)^2 \), and \( 2x^3 \) from \( (2x)^3 \)
Distribute a negative sign across a grouped expression: \( -(a+b) = -a - b \)
Read complex rational expressions like \( \dfrac{1}{x+h} - \dfrac{1}{x} \) without ambiguity
Section 2: Prerequisites
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This lesson follows COM-1. The notation habits you established there — vertical work, exact values, clear equals signs — are expected throughout. You'll also need:
From COM-1:
Vertical work: Each algebraic step on its own line — this matters especially when untangling BEDMAS chains
Exact values: Leave answers as fractions, not decimals (e.g., \( \frac12 \) not 0.5)
From before this course:
Integers and fractions: Arithmetic with positive and negative numbers, including multiplication of negatives
Basic exponents: That \( 3^2 = 9 \), \( 2^3 = 8 \) — the mechanics of repeated multiplication
Parentheses: That parentheses group operations and must be resolved before anything outside them
Before continuing, check your comfort level:
If negative arithmetic feels shaky, pause and review that first — it's the foundation for exponent binding and the minus sign distribution rules in Section 3.
Section 3: Core Concepts
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C1 — BEDMAS / PEMDAS: The Precedence Hierarchy
Mathematical notation is ambiguous without a fixed set of rules for which operations happen first. The solution is a precedence hierarchy: a ranked list of operations from "always first" to "always last."
BEDMAS / PEMDAS — Order of Operations
Evaluate operations in this order, from highest to lowest priority:
"Left to right" only applies within the same tier. Many students think "left to right" is the main rule and evaluate everything in the order they encounter it. The tower above is the real rule. \( 5 + 3 \times 2 \) is not \( (5+3) \times 2 = 16 \) — it's \( 5 + (3 \times 2) = 11 \), because multiplication outranks addition.
C2 — The Fraction Bar as a Grouping Symbol
The fraction bar is invisible parentheses. It groups everything in the numerator into one quantity and everything in the denominator into another. This means you must evaluate the full numerator and the full denominator before dividing.
The Fraction Bar
\( \dfrac{A}{B} \) means \( (A) \div (B) \). Whatever is written above the bar is one grouped expression; whatever is below is another. The division happens after both are evaluated.
In particular: \( \dfrac{x + 1}{x - 2} \neq x + \dfrac{1}{x} - 2 \). The first expression divides the whole quantity \( (x+1) \) by \( (x-2) \). The second is three separate terms.
This becomes critical when you read expressions typed in a single line. The expression:
(x + 1) / (x - 2)
is the correct typed version of the fraction. Without the parentheses, x + 1 / x - 2 means something completely different — only the \( 1 \) is divided by \( x \).
Typed fractions need explicit parentheses. When using a calculator or typing code, \( \dfrac{3x+5}{x-1} \) must be entered as (3x+5)/(x-1). Writing 3x+5/x-1 gives \( 3x + \frac{5}{x} - 1 \) — a completely different expression.
C3 — Nested Grouping: Work from the Inside Out
When grouping symbols are nested — parentheses inside brackets, brackets inside braces — you always start with the innermost grouping and work outward.
Nested Grouping
In an expression like \( 2 \times [3 \times (4 + 1)] \), the innermost group is \( (4 + 1) \). Evaluate it first, then the bracket, then the outer multiplication.
Typographically, mathematicians use different grouping symbols — \( () \), \( [] \), \( \{\} \) — purely to help the eye distinguish levels of nesting. They all mean the same thing: evaluate the contents before using the result.
Don't start with the outer grouping. \( 2 \times [3 \times (4+1)] \) — some students expand the bracket first: \( [6 \times (4+1)] = 6 \times 5 = 30 \). Coincidentally, this gives the same answer here (because multiplication is associative), but in more complex expressions the approach fails. Always start inside and work out.
C4 — Exponent Binding: What Does the Exponent Grab?
An exponent applies to the single token immediately to its left — nothing more. That token is called the base. If you want multiple things to form the base, you must group them with parentheses first.
Exponent Binding Rule
The base of an exponent is the single number, variable, or parenthesized group directly to the left of the exponent symbol.
\( -x^2 \): the base is \( x \), not \( -x \). So \( -x^2 = -(x^2) \).
\( (-x)^2 \): the base is \( (-x) \) — the parentheses make the negative part of the base.
\( 2x^3 \): the base is \( x \). So \( 2x^3 = 2 \cdot (x^3) \), not \( (2x)^3 \).
\( (2x)^3 \): the base is \( (2x) \). So \( (2x)^3 = 8x^3 \).
Let's make this concrete. At \( x = 3 \):
Four expressions, four different values:
\( -x^2 \) at \( x = 3 \): \( -(3^2) = -9 \)
\( (-x)^2 \) at \( x = 3 \): \( (-3)^2 = 9 \)
\( 2x^3 \) at \( x = 3 \): \( 2 \cdot (3^3) = 2 \cdot 27 = 54 \)
\( (2x)^3 \) at \( x = 3 \): \( (2 \cdot 3)^3 = 6^3 = 216 \)
The expression \( -3^2 \) that opened this lesson follows this rule: the base of the exponent is \( 3 \) (not \( -3 \)), so \( -3^2 = -(3^2) = -9 \). Every time you see a bare negative sign in front of an exponent — with no parentheses — the negative is outside the exponentiation.
C5 — Distributing vs. Not: The Reach of the Minus Sign
The minus sign is a versatile character in algebra. When it stands in front of a grouped expression, it distributes to every term inside. When it stands in front of a single term, it applies only to that term.
Distributing a Negative Sign
For any expression \( (A + B) \) or \( (A - B) \):
\[ -(A + B) = -A - B \qquad \text{(the minus flips every sign inside)} \]
\[ -(A - B) = -A + B \qquad \text{(the minus of a negative becomes positive)} \]
This follows from multiplying by \( -1 \): \( -(A+B) = (-1)(A+B) = (-1)(A) + (-1)(B) = -A - B \).
A very common mistake in calculus: when subtracting two expressions, students forget to flip all the signs in the subtracted expression.
Wrong: \( (5x^2 + 3x) - (2x^2 - x) = 5x^2 + 3x - 2x^2 - x \) (forgot to flip the sign on \( -x \))
Correct: \( (5x^2 + 3x) - (2x^2 - x) = 5x^2 + 3x - 2x^2 + x = 3x^2 + 4x \) (the minus distributes: \( -(\!-x) = +x \))
In calculus, you will compute the difference quotient \( f(x+h) - f(x) \) constantly. When \( f(x) \) has multiple terms — say \( f(x) = x^2 - 3x + 1 \) — the subtraction must distribute over every term of \( f(x) \): the result is \( \ldots - (x^2 - 3x + 1) = \ldots - x^2 + 3x - 1 \). Missing a sign flip here invalidates the entire derivative calculation.
C6 — Reading Complex Rational Expressions Without Ambiguity
As expressions become more involved, it becomes easy to mis-parse them — especially when multiple fraction bars or layered operations appear. The key is to identify what each fraction bar groups before doing any arithmetic.
Each Fraction Bar Has Its Own Scope
A fraction bar only covers what is directly written above and below it. It does not extend to absorb a term that is written to its right with a \( + \) or \( - \) sign.
\( \dfrac{1}{x+h} - \dfrac{1}{x} \) is a difference of two fractions. The two fraction bars are separate; neither absorbs the other.
\( \dfrac{1}{x+h - \frac{1}{x}} \) is a compound fraction: a single fraction with a complex denominator. Completely different expression.
The first expression — the difference of two fractions — appears at the heart of the difference quotient in calculus, which you'll compute in BTC-2. Misreading it as the second expression produces nonsensical results.
Always ask: what does each fraction bar span? Identify the numerator and denominator of each fraction separately before reading the overall expression.
Danger in typed expressions: When algebra is typed linearly (as in a calculator or code), the expression \( \dfrac{1}{x+h} - \dfrac{1}{x} \) becomes 1/(x+h) - 1/x. The subtraction sign is between two fractions — it does not extend the denominator of the first fraction. Reading it as 1/(x+h-1/x) is a fundamental misparse.
The key decision at Step 4: Why multiply before dividing? Because they're at the same level — and at the same level, you go left to right. \( 3 \times 18 \) appears before \( \div 6 \), so multiplication happens first.
Example 2 — The Fraction Bar as Grouping (Partially Scaffolded)
Evaluate \( \dfrac{2x + 4}{x - 1} \) at \( x = 5 \), and explain why \( 2x + \dfrac{4}{x} - 1 \) gives a different result.
Step 1 — Identify the structure: The fraction bar groups \( 2x + 4 \) as the entire numerator and \( x - 1 \) as the entire denominator. Think of it as \( (2x + 4) \div (x - 1) \).
Before continuing: what do you substitute in the numerator? What about the denominator?
\( \dfrac{7}{2} \neq \dfrac{49}{5} \). The fraction bar changes everything.
Example 3 — Exponent Binding and Sign Distribution (Minimally Scaffolded)
Simplify \( -(3x + 2)^2 \) for \( x = 1 \). Then compare it to \( (-3x - 2)^2 \) at \( x = 1 \).
Hint: Identify the base of each exponent, then handle the negative sign only after squaring.
Show Solution
For \( x = 1 \):
\( -(3x + 2)^2 \): the base of the exponent is \( (3x+2) \). The leading minus is outside the exponent.
\[ -(3(1)+2)^2 = -(3+2)^2 = -(5)^2 = -25 \]
\( (-3x - 2)^2 \): the entire expression \( (-3x-2) \) is the base. Squaring a negative gives positive.
\[ (-3(1)-2)^2 = (-3-2)^2 = (-5)^2 = 25 \]
Result: \( -25 \neq 25 \). One minus sign, placed differently, changes the result by a factor of \( -1 \).
Also: Note that \( (-3x - 2) = -(3x + 2) \), so \( (-3x-2)^2 = (-(3x+2))^2 = (3x+2)^2 \). Squaring eliminates the negative. But \( -(3x+2)^2 \) never squares the negative — it negates the result of squaring the positive quantity.
Example 4 — Reading a Complex Rational Expression (Application Twist)
This is the kind of expression you'll meet in calculus. Read it carefully before touching any numbers.
Problem: What is the correct reading of \( \dfrac{1}{x+h} - \dfrac{1}{x} \)? Compute its value for \( x = 4, \; h = 1 \).
Challenge yourself first: a student claims this equals \( \dfrac{1}{x + h - \frac{1}{x}} \). Is the student right? Why or why not?
Show Solution
The expression \( \dfrac{1}{x+h} - \dfrac{1}{x} \) is a difference of two separate fractions. The minus sign is between them — it does not merge them under one bar.
The student is wrong. Extending the denominator of the first fraction to absorb \( - \dfrac{1}{x} \) is a misparse. The two fraction bars are independent.
Finding a common denominator: \( \dfrac{4}{20} - \dfrac{5}{20} = -\dfrac{1}{20} \)
What the student's expression gives: \( \dfrac{1}{x + h - \frac{1}{x}} = \dfrac{1}{4 + 1 - \frac{1}{4}} = \dfrac{1}{5 - 0.25} = \dfrac{1}{4.75} \approx 0.21 \)
\( -0.05 \neq 0.21 \). The two expressions are completely different.
Section 5: Guided Practice
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Choose the best answer from each dropdown. Incorrect choices will explain exactly why they're wrong — read those explanations; they target the most common traps.
Problem 1 — BEDMAS Order (C1)
A student evaluates \( 5 + 3 \times 2 \) by computing \( 5 + 3 = 8 \), then \( 8 \times 2 = 16 \). What is the error?
A student evaluates \( 24 \div 6 - 2 \) by computing \( 24 \div (6 - 2) = 24 \div 4 = 6 \). What went wrong?
A student evaluates \( 2 + 3^2 \) by computing \( 2 + 3 = 5 \), then \( 5^2 = 25 \). What is the error?
A student evaluates \( 3 \times 4 + 2 \) by first computing \( 4 + 2 = 6 \), then \( 3 \times 6 = 18 \). What is the error?
A student evaluates \( 15 \div 3 \times 5 \) by first computing \( 3 \times 5 = 15 \), then \( 15 \div 15 = 1 \). What went wrong?
Problem 2 — The Fraction Bar as Grouping (C2)
Evaluate \( \dfrac{x + 5}{2} \) at \( x = 3 \).
Which expression correctly encodes "the sum of \( a \) and \( b \), divided by 3"?
Evaluate \( \dfrac{8 - 2}{3} \).
Evaluate \( \dfrac{2y + 6}{4} \) at \( y = 1 \).
Which expression correctly encodes "the difference of \( p \) and \( q \), divided by 2"?
Problem 3 — Exponent Binding (C4)
Which of the following statements is true?
Problem 4 — Distributing the Minus Sign (C5)
A student simplifies \( -(3x - 5) \) as \( -3x - 5 \). What went wrong?
Section 6: Independent Practice
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These problems are unscaffolded. Work through each one, then reveal the solution to check your reasoning — not just your answer.
Problem 1 — Evaluate with BEDMAS (C1, C3) — regenerable
Problem 2 — Fraction Bar Grouping (C2) — regenerable
Problem 3 — Exponent Binding: Identify the Error (C4)
A student evaluates \( -5^2 \) by computing \( (-5)^2 = 25 \). Which statement correctly explains the error?
Which of the following equals \( 2x^3 \) when \( x = 3 \)?
A student claims \( -x^2 = (-x)^2 \) "because a negative times a negative is positive." What is wrong with this reasoning?
A student evaluates \( 3 \cdot 2^2 \) by computing \( (3 \cdot 2)^2 = 6^2 = 36 \). Which statement correctly explains the error?
A student says \( -(4^2) \) and \( (-4)^2 \) are both equal to 16 because "squaring always makes the result positive." What is wrong?
Problem 4 — Verbal Description to Expression (C2, C5)
Which expression correctly encodes "5 times the sum of \( x \) and 2, minus 3"?
Which expression correctly encodes "\( x \) plus 4, all divided by the difference of \( x \) and 3"?
Which expression correctly encodes "the square of the sum of \( a \) and \( b \), plus \( c \)"?
Which expression correctly encodes "3 less than twice \( x \)"?
Which expression correctly encodes "the cube of the difference of \( x \) and 1"?
Problem 5 — Complex Rational Structure (C6)
Consider the expression \( \dfrac{1}{x+h} - \dfrac{1}{x} \). A student "simplifies" it by writing \( \dfrac{1}{x + h - \frac{1}{x}} \), claiming to have "combined the terms into one fraction." What is the fundamental error?
Section 7: Mastery Check
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No hints. No scaffolding. These questions test whether the concepts have actually landed.
Feynman Test — Explain the Fraction Bar Rule (C2, C3)
A classmate just got this wrong on a quiz: they evaluated \( \dfrac{2x + 6}{2} \) as \( 2x + 3 \) instead of \( x + 3 \). Explain, in your own words, what the fraction bar rule is and why it changes the answer. Mention at least one example.
Characters: 0/600
See a model explanation
The fraction bar acts like invisible parentheses around the entire numerator and the entire denominator. So \( \dfrac{2x+6}{2} \) means \( (2x+6) \div 2 \) — every term in the numerator is divided by 2, not just the last one. Factoring: \( \dfrac{2x+6}{2} = \dfrac{2(x+3)}{2} = x + 3 \). The classmate's error was dividing only the 6 by 2 (getting \( 2x + 3 \)) instead of dividing the whole numerator. Another example: \( \dfrac{3+9}{3} = \dfrac{12}{3} = 4 \), not \( 3 + 3 = 6 \).
Apply — Multi-Layer BEDMAS (C1, C4)
In the expression \( -2 \cdot (3 + 4^2) \div (-1)^3 \), which step must be performed first?
A student evaluates \( 12 - 3^2 \) with the following work:
Line 1: \( 12 - 3^2 \)
Line 2: \( = (12 - 3)^2 \)
Line 3: \( = 9^2 \)
Line 4: \( = 81 \)
The correct answer is \( 12 - 9 = 3 \). Identify the error and state which rule it violates.
Show Explanation
Error in Line 2: The student moved the subtraction inside the exponent by grouping \( (12-3)^2 \). This violates the exponent binding rule (C4): exponents bind to the single token immediately to their left.
In \( 12 - 3^2 \), the exponent applies only to \( 3 \). The correct first step is to compute \( 3^2 = 9 \) (exponents before subtraction), giving \( 12 - 9 = 3 \).
There are no parentheses around \( (12-3) \) in the original expression. The student invented a grouping that does not exist.
Self-Assessment
How confident are you with the four core concepts from this lesson?
Still confusedVery confident
Section 8: Boss Fight
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You've practiced individual concepts. Now it's time to use all four at once — under pressure. Choose your path:
🔍 The Debugger
You receive a student's flawed solution. Your job: find every error, name the rule it breaks, and rewrite the solution correctly. Think like a professor with a red pen.
🏗️ The Encoder
You receive verbal descriptions of calculations. Your job: translate each one into a precise algebraic expression and evaluate it. One misplaced parenthesis changes everything.
Path A: The Debugger 🔍
Mélodie is evaluating \( -3^2 + 2 \times (4 + 1) \). She shows her work step by step. Two errors are embedded — one violates C4 (exponent binding) and one violates C1 (order of operations).
Identify Error 1: which step, which concept (C1–C6), and what Mélodie did wrong.
Identify Error 2: which step, which concept, and what Mélodie did wrong.
Rewrite the full solution correctly, with vertical work.
Show Solution
Error 1 — Step 2 (C4: Exponent Binding):
Mélodie rewrote \( -3^2 \) as \( (-3)^2 \), claiming they are "the same thing." They are not. In \( -3^2 \), the exponent binds to \( 3 \) only — the negative is outside the exponent. So \( -3^2 = -(3^2) = -9 \). In \( (-3)^2 \), the negative is part of the base, giving \( (-3)(-3) = 9 \). These are different expressions with different values.
Error 2 — Step 4 (C1: Order of Operations):
Mélodie added \( 9 + 2 = 11 \) before resolving the multiplication \( 2 \times (\ldots) \). Multiplication has higher precedence than addition. The \( 2 \) must multiply the bracket result before the \( 9 \) (or rather \( -9 \), after fixing Error 1) is added.
Reflection: Error 1 is the sneaky one — Mélodie's notation looked plausible. When a negative sign appears before an exponent with no parentheses, always read it as "negate the result of the exponent," not as "include the negative in the base."
Path B: The Encoder 🏗️
Three verbal descriptions of calculations are given below. For each one, your job is to: (1) write the correct algebraic expression — with all necessary parentheses — and (2) evaluate it at \( x = 3 \).
Be precise: a missing parenthesis changes the meaning and the value.
Description 1: "5 minus x squared"
Description 2: "The sum of \( x \) and 2, all divided by \( x \) minus 1"
Description 3: "The opposite of 2 times the sum of \( x \) and 4"
Common error: Writing \( (5-x)^2 \) instead. "5 minus x squared" squares only \( x \); the subtraction comes after. If the intent were to square the difference, the description would say "the square of the quantity 5 minus x."
Description 2: "The sum of \( x \) and 2, all divided by \( x \) minus 1"
Expression: \( \dfrac{x+2}{x-1} \) or equivalently \( (x+2)/(x-1) \) in typed form
At \( x = 3 \): \( \dfrac{3+2}{3-1} = \dfrac{5}{2} \)
Common error: Writing \( x + 2/x - 1 \), which means \( x + \frac{2}{x} - 1 = 3 + \frac{2}{3} - 1 = 2 + \frac{2}{3} = \frac{8}{3} \). The phrase "all divided by" signals that the entire sum \( (x+2) \) is the numerator.
Description 3: "The opposite of 2 times the sum of \( x \) and 4"
Common error 1: Writing \( -2x + 4 \). This reads as "multiply \( x \) by \(-2\) then add 4" — not "multiply the sum \( (x+4) \) by \(-2\)." At \( x=3 \): \( -2(3) + 4 = -2 \neq -14 \).
Common error 2: Writing \( -2x - 4 \) (distributing the negative but also negating the constant). Check: \( -2(x+4) = -2x - 8 \), not \( -2x - 4 \).
Reflection: Verbal descriptions can be ambiguous — that's why mathematical notation was invented. The goal is to translate precisely enough that the algebraic expression has exactly one interpretation.
Section 9: Challenge Problems
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Optional stretch. These go beyond the lesson objectives. Attempt them when you feel confident with the core material.
Evaluate \( \dfrac{1}{1 + \dfrac{1}{x}} \) at \( x = 2 \). Then simplify the expression algebraically for general \( x \), writing the result as a single fraction.
Algebraic simplification:
\( \dfrac{1}{1 + \frac{1}{x}} \) — the denominator is \( 1 + \dfrac{1}{x} = \dfrac{x}{x} + \dfrac{1}{x} = \dfrac{x+1}{x} \)
So the expression equals \( \dfrac{1}{\frac{x+1}{x}} = 1 \cdot \dfrac{x}{x+1} = \dfrac{x}{x+1} \)
Check at \( x = 2 \): \( \dfrac23 = \dfrac23 \) ✓
This is a compound fraction — a fraction whose denominator itself contains a fraction. The key move is to simplify the inner denominator first (C3: innermost group first), then divide.
Challenge 2 — Exponent Binding Extension (C4)
Without a calculator, determine which is larger: \( -2^{10} \) or \( (-2)^{10} \). Explain why in one sentence. Then determine which is larger: \( -2^{11} \) or \( (-2)^{11} \).
Show Solution
\( -2^{10} = -(2^{10}) = -1024 \). The negative is outside the exponent.
\( (-2)^{10} = 2^{10} = 1024 \). The negative is inside, and squaring (or any even power) eliminates it.
\( (-2)^{10} = 1024 > -1024 = -2^{10} \). Reason: \( -2^{10} \) is negative; \( (-2)^{10} \) is positive because an even power of a negative number is positive.
\( -2^{11} = -(2^{11}) = -2048 \).
\( (-2)^{11} = -(2^{11}) = -2048 \). An odd power of a negative number is negative.
So \( -2^{11} = (-2)^{11} = -2048 \). They are equal.
This is a preview of what you'll see in ALG-2: the rules for exponents interact with negative signs in non-obvious ways, and keeping track of what the base actually is matters enormously.
Challenge 3 — Preview: Setting Up for Exponent Laws (C2, C4)
Rewrite \( \dfrac{x^3 - x}{x^2} \) by splitting it into separate fractions, then simplify each one. What rule about exponents are you implicitly using?
Show Solution
Split using the fraction bar's scope:
\[ \frac{x^3 - x}{x^2} = \frac{x^3}{x^2} - \frac{x}{x^2} \]
Simplify each fraction:
\( \dfrac{x^3}{x^2} = x^{3-2} = x^1 = x \) (subtracting exponents when dividing)
\( \dfrac{x}{x^2} = x^{1-2} = x^{-1} = \dfrac{1}{x} \) (negative exponent means reciprocal)
Result: \( x - \dfrac{1}{x} \)
The implicit rule: when dividing powers of the same base, you subtract exponents — \( x^a / x^b = x^{a-b} \). This is one of the fundamental exponent laws you'll develop formally in ALG-2. The fraction bar made the structure visible enough to apply it.
Section 10: Solutions Reference
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Complete, step-by-step solutions for all problems in Sections 5–9 are available on the solutions page. Solutions include common mistakes and what to check if your answer doesn't match.
What to do if you're stuck: Re-read the relevant Core Concept in Section 3, then try the worked example that corresponds to the concept (the example numbers align with the concept numbers). The solutions page shows the reasoning behind every step, not just the answer.