ALG-1 Solutions: Order of Operations and Algebraic Structure

Problem 1 — BEDMAS Order (C1)

Variant 0 — Student evaluates \( 5 + 3 \times 2 \) by adding first: answer is B.

Multiplication has higher precedence than addition. \( 3 \times 2 = 6 \) must be computed before adding 5: \( 5 + 6 = 11 \), not 16.

• A (arithmetic wrong): The arithmetic \( 5+3=8 \) is correct; the issue is which operation comes first.
• C (both valid): Precedence is not optional. There is exactly one correct answer: 11.
• D (should have used parens): Without parentheses, BEDMAS applies. If you write \( (5+3) \times 2 \), then 16 would be correct — but the original has no parentheses.

Variant 1 — Student evaluates \( 24 \div 6 - 2 \) by grouping the subtraction: answer is C.

Without parentheses, division comes before subtraction. \( 24 \div 6 = 4 \) first, then \( 4 - 2 = 2 \). The student mentally inserted parentheses that don't exist.

Variant 2 — Student evaluates \( 2 + 3^2 \) by adding first: answer is B.

Exponents come before addition. \( 3^2 = 9 \) first, then \( 2 + 9 = 11 \). The student incorrectly treated \( 2 + 3 \) as the base of the exponent.

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Problem 2 — Fraction Bar as Grouping (C2)

Variant 0 — Evaluate \( \dfrac{x+5}{2} \) at \( x = 3 \): answer is B (value = 4).

The fraction bar groups the entire numerator: \( \dfrac{3+5}{2} = \dfrac{8}{2} = 4 \). The option "5.5" misreads the expression as \( 3 + \frac{5}{2} \).

Variant 1 — "Sum of \( a \) and \( b \) divided by 3": answer is B — (a + b) / 3.

The word "sum" signals that \( a + b \) is the complete numerator. Only option B groups it correctly. \( a + b/3 \) divides only \( b \); \( a/3 + b \) divides only \( a \).

Variant 2 — Evaluate \( \dfrac{8-2}{3} \): answer is B (value = 2).

The fraction bar groups \( 8-2 = 6 \) as the numerator. Then \( 6 \div 3 = 2 \). The distractor "7⅓" misreads the expression as \( 8 - \frac{2}{3} \).

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Problem 3 — Exponent Binding (C4)

Answer is C.

In \( -4^2 \), the exponent binds to \( 4 \) only: \( -4^2 = -(4^2) = -16 \). In \( (-4)^2 \), the negative is part of the base: \( (-4)(-4) = 16 \). They have opposite signs. Additionally: \( 3 \cdot 4^2 = 3 \cdot 16 = 48 \), while \( (3 \cdot 4)^2 = 12^2 = 144 \). All four expressions are distinct.

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Problem 4 — Distributing the Minus Sign (C5)

Answer is A.

\( -(3x - 5) = (-1)(3x - 5) = (-1)(3x) + (-1)(-5) = -3x + 5 \). The student negated \( 3x \) correctly but forgot that \( -(-5) = +5 \), not \( -5 \). Every term inside the parentheses changes sign.

Problem 1 — Evaluate with BEDMAS (Generator)

This problem is parameterized — it generates a new numerical expression each time. Three expression templates are used:

Template 1: \( a + b \times c^2 \)

Method: exponent first (\( c^2 \)), then multiplication (\( b \times c^2 \)), then addition. The key error to avoid is computing \( a + b \) before multiplying.

Template 2: \( a \times (b + c) - d \)

Method: bracket first (\( b + c \)), then multiplication (\( a \times \text{result} \)), then subtraction. The key error to avoid is subtracting \( d \) before multiplying.

Template 3: \( (a + b)^2 - c \times d \)

Method: bracket first (\( a+b \)), then exponent (square the sum), then multiplication (\( c \times d \)), then subtraction. The key error to avoid is subtracting before multiplying, or squaring \( a \) and \( b \) separately.

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Problem 2 — Fraction Bar Grouping (Generator)

This problem generates a fraction \( \dfrac{a+b}{c} \) and asks you to contrast the correct value with a common misreading.

General method: The fraction bar groups the entire numerator \( (a+b) \). Evaluate the numerator completely, then divide by \( c \). The wrong reading \( a + b/c \) divides only \( b \) by \( c \) — this ignores the fraction bar's scope.

Correct: \( \dfrac{a+b}{c} = \dfrac{a+b}{c} \) — numerator evaluated first.
Wrong: \( a + \dfrac{b}{c} \) — only the last term is divided.

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Problem 3 — Exponent Binding: Identify the Error (C4)

Variant 0 — Student evaluates \( -5^2 \) as \( (-5)^2 = 25 \): answer is B.

\( -5^2 = -(5^2) = -25 \). The exponent applies to \( 5 \) alone. Writing \( (-5)^2 \) puts the negative inside the base — a structurally different expression.

Variant 1 — Which equals \( 2x^3 \) at \( x = 3 \): answer is B (54).

\( 2x^3 = 2 \cdot (x^3) = 2 \cdot 27 = 54 \). Option A computes \( (2x)^3 = 6^3 = 216 \) — that is a different expression where the coefficient 2 is inside the base.

Variant 2 — Why \( -x^2 \neq (-x)^2 \): answer is C.

In \( -x^2 \), the exponent binds to \( x \) only: \( -x^2 = -(x^2) \). This is negative for all \( x \neq 0 \). In \( (-x)^2 \), the whole \( (-x) \) is the base, and squaring a negative gives positive. The "negative times negative" reasoning is valid for \( (-x)^2 \) but misapplied to \( -x^2 \).

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Problem 4 — Verbal Description to Expression (C2, C5)

Variant 0 — "5 times the sum of \( x \) and 2, minus 3": answer is B — 5(x + 2) − 3.

"5 times the sum" means \( 5 \) multiplies the entire sum \( (x+2) \). The "minus 3" is outside the multiplication. \( 5x + 2 - 3 \) only multiplies \( x \) by 5; \( 5(x+2-3) \) subtracts 3 before multiplying.

Variant 1 — "\( x \) plus 4, all divided by \( x \) minus 3": answer is B — (x + 4) / (x − 3).

"All divided by" signals that both \( (x+4) \) and \( (x-3) \) must be grouped. Any option that fails to parenthesize both the numerator and denominator is wrong.

Variant 2 — "Square of the sum of \( a \) and \( b \), plus \( c \)": answer is B — (a + b)² + c.

"Square of the sum" means \( (a+b) \) is the base. The \( + c \) is added after squaring. \( a^2 + b^2 + c \) applies the square to each variable separately (wrong). \( (a+b+c)^2 \) also squares \( c \) (wrong).

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Problem 5 — Complex Rational Structure (C6)

Answer is B.

\( \dfrac{1}{x+h} - \dfrac{1}{x} \) is a difference of two fractions. Each fraction bar is independent; neither extends to the other. Combining them into one fraction requires finding a common denominator:

\[ \frac{1}{x+h} - \frac{1}{x} = \frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)} \]

The student's expression \( \dfrac{1}{x+h-\frac{1}{x}} \) has a completely different structure — the denominator of a single fraction, not a difference.

Feynman Test — Model Explanation

The fraction bar acts like invisible parentheses around the entire numerator and the entire denominator. So \( \dfrac{2x+6}{2} \) means \( (2x+6) \div 2 \), which divides every term by 2: \( \dfrac{2x}{2} + \dfrac{6}{2} = x + 3 \). The classmate's error was treating only 6 as the numerator, computing \( 2x + \frac{6}{2} = 2x + 3 \) instead. A simple test: try \( x = 0 \). Correct: \( \frac{0+6}{2} = 3 \). Classmate's version: \( 0 + 3 = 3 \). Coincidentally equal at \( x = 0 \). Try \( x = 2 \). Correct: \( \frac{4+6}{2} = 5 \). Classmate's version: \( 4 + 3 = 7 \). Now the error shows.

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Apply — Multi-Layer BEDMAS

First step: answer is B — compute 4² = 16 (exponent inside the bracket).

Full solution:

\[ -2 \cdot (3 + 4^2) \div (-1)^3 = -2 \cdot (3 + 16) \div (-1) = -2 \cdot 19 \div (-1) = -38 \div (-1) = 38 \]

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Analyze — Error in \( 12 - 3^2 \)

Error location: Line 2 — the student wrote \( (12 - 3)^2 \), inventing grouping parentheses that don't exist in the original expression.

Concept violated: C4 (Exponent Binding). The exponent in \( 12 - 3^2 \) applies only to \( 3 \).

Correct solution: \( 12 - 3^2 = 12 - 9 = 3 \)

Path A: The Debugger

Error 1 — Step 2 (C4: Exponent Binding):

Mélodie rewrote \( -3^2 \) as \( (-3)^2 \). These are not the same. \( -3^2 = -(3^2) = -9 \). Moving the negative inside the exponent changes the expression from \(-9\) to \(+9\).

Error 2 — Step 4 (C1: Order of Operations):

Mélodie computed \( 9 + 2 = 11 \) before resolving the multiplication \( 2 \times (4+1) \). Multiplication must come before addition: evaluate the bracket \( (4+1) = 5 \) first, then multiply \( 2 \times 5 = 10 \), then add \( -9 + 10 = 1 \).

Correct solution:

\[ -3^2 + 2 \times (4 + 1) = -9 + 2 \times 5 = -9 + 10 = 1 \]

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Path B: The Encoder

Description 1: "5 minus x squared" → \( 5 - x^2 \)

At \( x = 3 \): \( 5 - 3^2 = 5 - 9 = -4 \).
Not \( (5-x)^2 = 4 \). "Minus x squared" means subtract \( x^2 \), not square the difference.

Description 2: "Sum of \( x \) and 2, all divided by \( x \) minus 1" → \( \dfrac{x+2}{x-1} \)

At \( x = 3 \): \( \dfrac{3+2}{3-1} = \dfrac{5}{2} \).
Not \( x + 2/x - 1 = 3 + \frac{2}{3} - 1 = \frac{8}{3} \). "All divided by" signals the entire sum is the numerator.

Description 3: "Opposite of 2 times the sum of \( x \) and 4" → \( -2(x+4) \)

At \( x = 3 \): \( -2(3+4) = -2(7) = -14 \).
Not \( -2x + 4 = -2 \). "Times the sum" means 2 multiplies the whole \( (x+4) \).

Challenge 1 — Nested Compound Fraction

Evaluate \( \dfrac{1}{1 + \frac{1}{x}} \) at \( x = 2 \): answer is \( \dfrac{2}{3} \).

General form: simplify the denominator \( 1 + \frac{1}{x} = \frac{x+1}{x} \), then divide: \( \dfrac{1}{\frac{x+1}{x}} = \dfrac{x}{x+1} \).

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Challenge 2 — Exponent Binding Extension

\( (-2)^{10} = 2^{10} = 1024 > -1024 = -2^{10} \). Even power makes the negative disappear.

\( -2^{11} = (-2)^{11} = -2048 \). Odd power preserves the negative — both expressions equal \( -2048 \).

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Challenge 3 — Preview: Exponent Laws

\( \dfrac{x^3 - x}{x^2} = \dfrac{x^3}{x^2} - \dfrac{x}{x^2} = x - x^{-1} = x - \dfrac{1}{x} \)

Implicit rule: \( x^a / x^b = x^{a-b} \) (subtract exponents when dividing same base). Fully developed in ALG-2.