Foundations Reteach Guide Indiana Academic Standards • Grade 8 Mathematics
Exponents & Linear Equations Guide
Grade 8 Targets Standards 8.NS.3 & 8.AF.1
Review these core properties, algorithms, and worked examples before beginning the independent practice.
Standard 8.NS.3
Properties of Integer Exponents
Indiana Academic Standard
Given a numeric expression with common rational number bases and integer exponents, apply the properties of exponents to generate equivalent expressions.
Product Rule \(a^m \cdot a^n = a^{m+n}\)
Quotient Rule \(\frac{a^m}{a^n} = a^{m-n}\)
Power of Power \((a^m)^n = a^{m \cdot n}\)
Negative / Zero \(a^{-n} = \frac{1}{a^n}\), \(a^0 = 1\)
Worked Model Problem Simplify to a single power: \(\frac{(4^3 \cdot 4^{-5})^2}{4^{-8}}\)
Step 1: Parentheses First
Apply product rule inside:
Step 2: Power to a Power
Multiply exponents:
Step 3: Quotient Rule
Subtract bottom from top:Final Answer: \(4^4\) (or \(256\))
Watch Out: \(4^{-2} \neq -8\) or \(-16\). Negative exponents mean reciprocals: \(4^{-2} = \frac{1}{4^2} = \frac{1}{16}\).
Standard 8.AF.1
Multi-Step Linear Equations
Indiana Academic Standard
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
Phase 1: Simplify Each Side
Clear parentheses using the distributive property. Combine like terms that are on the same side of the equals sign.
Phase 2: Collect & Isolate
Use inverse operations to collect all variable terms on one side, then isolate the variable by undoing additions/subtractions, then multiplying/dividing.
Worked Model Problem Solve: \(3(2x - 4) + 5 = 2x - 15\)
Step 1: Distribute & Combine
\(6x - 12 + 5 = 2x - 15\)
Step 2: Collect Variable Terms
Subtract \(2x\) from both sides:
Step 3: Isolate Variable
Add \(7\): \(4x = -8\)Solution: \(x = -2\)
Watch Out: Do not add \(6x\) and \(2x\) across the equals sign. To move a term across \(=\), use inverse operations!
Indiana Academic Standards Reteach Guide • Grade 8 Page 1 of 2
Indiana Academic Standards • Grade 8 Mathematics
Classifying Linear Equation Solutions
Grade 8 Target Standard 8.AF.2
Distinguishing between one solution, no solution, and infinitely many solutions.
Standard 8.AF.2
Number of Solutions Classification
Indiana Academic Standard
Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which is the case by successively transforming equations into simpler forms (\(x = a\), \(a = a\), or \(a = b\)).
1 One Unique Solution
Simplifies to: \(x = a\) Feature: Variable coefficients are different on each side.Example: \(3x + 2 = 5x - 4 \implies x = 3\)
∅ No Solution
Simplifies to: \(a = b\) (False statement)Feature: Same coefficient of \(x\), but different constants .Example: \(2x + 5 = 2x - 3 \implies 5 = -3\)
∞ Infinitely Many Solutions
Simplifies to: \(a = a\) (Identity / True)Feature: Same coefficient of \(x\) and identical constants .Example: \(4x + 6 = 2(2x + 3) \implies 6 = 6\)
Worked Model: Transforming & Proving Solution Types
Model A: Prove No Solution
\(4(x - 2) = 4x + 9\)
Distribute: \(4x - 8 = 4x + 9\)
Subtract \(4x\): \(-8 = 9\)
Conclusion: False statement → No Solution (∅)
Model B: Prove Infinitely Many
\(-2(3x - 5) + 4 = -6x + 14\)
Distribute: \(-6x + 10 + 4 = -6x + 14\)
Combine: \(-6x + 14 = -6x + 14\)
Add \(6x\): \(14 = 14\)
Conclusion: Identity → Infinitely Many (∞)
Targeting Solutions: What values make \(6x + 10 = \mathbf{A}x + \mathbf{B}\) have...
No Solution: Match coefficients, change constant!
Infinitely Many: Match both coefficient and constant!
Grade 8 Algebra Decision Protocol
1. Clear Exponents In numeric terms, apply product, quotient, and power rules first.
2. Expand & Combine Distribute all multipliers and combine like terms on each separate side.
3. Compare Terms Look at coefficients of \(x\). If equal, check constants for \(\∅\) or \(\∞\).
4. Conclude & Verify State the unique solution value or justify why no/infinite solutions exist.
Indiana Academic Standards Reteach Guide • Grade 8 Page 2 of 2
Foundations Practice Worksheet Indiana Academic Standards • Grade 8 Mathematics
Grade 8 Algebra Foundations
Page 1 of 2 (Front)
Standards 8.NS.3 & 8.AF.1
Student Name:
Date:
Period:
Self-Check:
/ 6 pts
SECTION 1
Indiana Academic Standard 8.NS.3
Properties of Exponents
Given a numeric expression with common rational number bases and integer exponents, apply the properties of exponents to generate equivalent expressions.
1. Product & Quotient Rules: Base 5
Simplify: \(\frac{5^4 \cdot 5^{-7}}{5^{-5}}\)
Single Power:
2. Power to a Power Rule: Mixed Bases
Simplify: \((2^3 \cdot 3^{-2})^2 \cdot 3^5\)
Simplified:
3. Scientific Application: Context
A computer processor completes \((2^4)^3\) calculations per microsecond. During a task, it divides those operations across \(2^{15}\) memory channels: \(\frac{(2^4)^3}{2^{15}}\). Express operations per channel as a fraction.
Final Value:
SECTION 2
Indiana Academic Standard 8.AF.1
Multi-Step Equations
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
4. Distribute & Collect: Solve for \(x\)
\(4(2x - 3) - 5 = 3x + 8\)
Solution:
5. Multi-Step with Negative: Solve for \(w\)
\(-3(2w + 4) = 2(w - 14)\)
Solution:
6. Gym Cost Comparison: Real Context
FitZone charges \(\$25/\text{month} + \$45\): \(25m + 45\). IronGym charges \(\$35/\text{month} + \$15\): \(35m + 15\). Write and solve an equation to find the months, \(m\), when both cost the same.
Months \(m\):
Algebra Foundations Reteach • Indiana Academic Standards 8.NS.3 & 8.AF.1 Turn over for Section 3 (8.AF.2) →
Indiana Academic Standards • Grade 8 Mathematics
Grade 8 Algebra Foundations
Page 2 of 2 (Back)
Standard 8.AF.2 & Error Analysis
Student Name: __________________________________________
Class Period: ___________
SECTION 3
Indiana Academic Standard 8.AF.2
Solution Classification
Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which is the case by successively transforming equations into simpler forms (\(x = a\), \(a = a\), or \(a = b\)).
7. Transform & Classify: Solve & State
\(5(2x - 4) = 10x - 12\)
Solution Type:
8. Transform & Classify: Solve & State
\(3(4x + 2) - 6x = 2(3x + 3)\)
Solution Type:
9. Solve & State Exact Value: Solve for \(y\)
\(2(3y - 1) = 4y + 10\)
Type:
Value of \(y\):
10. Create Target Equations: Find \(A\) & \(B\)
Given: \(8x - 14 = \mathbf{A}x + \mathbf{B}\) . Choose values of \(\mathbf{A}\) and \(\mathbf{B}\) for each condition below:
No Solution: \(A=\)___, \(B=\)___
Infinitely Many: \(A=\)___, \(B=\)___
Problem 11: Error Analysis & Exponent Misconception
Critique & Correct
Incorrect Student Claim:
A student simplifies: \(\frac{6^3}{6^{-2}}\) and writes:
"Quotient rule: subtract exponents."
\(\frac{6^3}{6^{-2}} = 6^{3 - 2} = 6^1 = 6\)
Explain the student's subtraction mistake:
Your Correct Step-by-Step Work:
Apply \(\frac{a^m}{a^n} = a^{m - n}\) accurately with negative integers.
Correct Value:
Algebra Foundations Reteach • Indiana Academic Standard 8.AF.2 Completed Practice Sheet • Check work with Answer Key
Foundations Solutions Key Teacher Resource • Grade 8 Worked Key
Grade 8 Algebra Solutions Key
Key: Page 1 of 2
Standards 8.NS.3 & 8.AF.1
SECTION 1 KEY
Indiana Academic Standard 8.NS.3: Properties of Exponents
Total: 3 Points
Problem 1 Solution 1 Point
\(\frac{5^4 \cdot 5^{-7}}{5^{-5}}\)
1. Numerator product rule: \(5^{4 + (-7)} = 5^{-3}\)
2. Quotient rule: \(\frac{5^{-3}}{5^{-5}} = 5^{-3 - (-5)}\)
3. Subtracting negative: \(5^{-3 + 5} = 5^2\)
4. Value: \(5^2 = 25\)
Correct Answer: \(5^2\) (or \(25\))
Error Note: Watch out for \(-3 - (-5) \neq -8\). Subtracting \(-5\) adds \(+5\).
Problem 2 Solution 1 Point
\((2^3 \cdot 3^{-2})^2 \cdot 3^5\)
1. Power to a power: \(2^{3 \cdot 2} \cdot 3^{-2 \cdot 2} = 2^6 \cdot 3^{-4}\)
2. Multiply with \(3^5\): \(2^6 \cdot (3^{-4} \cdot 3^5)\)
3. Combine base 3: \(3^{-4 + 5} = 3^1 = 3\)
4. Evaluate: \(2^6 \cdot 3 = 64 \cdot 3 = 192\)
Correct Answer: \(2^6 \cdot 3\) or \(192\)
Error Note: Students cannot combine base 2 and base 3 into base 6!
Problem 3 Solution (Processor Operations) 1 Point
Expression: \(\frac{(2^4)^3}{2^{15}}\)
Step 1: \((2^4)^3 = 2^{4 \cdot 3} = 2^{12}\)
Step 2: \(\frac{2^{12}}{2^{15}} = 2^{12 - 15} = 2^{-3}\)
Step 3: Convert negative exponent: \(2^{-3} = \frac{1}{2^3} = \frac{1}{8}\)
Correct Answer: \(\frac{1}{8}\) (or \(2^{-3}\)) (0.125 operations/channel)
SECTION 2 KEY
Indiana Academic Standard 8.AF.1: Multi-Step Linear Equations
Total: 3 Points
Problem 4 Solution 1 Point
\(4(2x - 3) - 5 = 3x + 8\)
1. Distribute: \(8x - 12 - 5 = 3x + 8\)
2. Combine left constants: \(8x - 17 = 3x + 8\)
3. Subtract \(3x\): \(5x - 17 = 8\)
4. Add 17: \(5x = 25 \implies x = 5\)
Correct Answer: \(x = 5\)
Problem 5 Solution 1 Point
\(-3(2w + 4) = 2(w - 14)\)
1. Distribute both sides: \(-6w - 12 = 2w - 28\)
2. Subtract \(2w\): \(-8w - 12 = -28\)
3. Add 12: \(-8w = -16\)
4. Divide by \(-8\): \(w = \frac{-16}{-8} = 2\)
Correct Answer: \(w = 2\)
Problem 6 Solution (Gym Memberships) 1 Point
Equation: \(25m + 45 = 35m + 15\)
Subtract \(25m\): \(45 = 10m + 15\)
Subtract 15: \(30 = 10m \implies m = 3\)
Check: FitZone \(= 25(3)+45 = \$120\); IronGym \(= 35(3)+15 = \$120\) ✓
Correct Answer: \(m = 3\text{ months}\) (Total Cost = $120)
Algebra Foundations Solutions Key • Standards 8.NS.3 & 8.AF.1 Page 1 of 2
Teacher Resource • Grade 8 Worked Key
Grade 8 Algebra Solutions Key
Key: Page 2 of 2
Standard 8.AF.2 & Error Analysis
Solutions for solution-type classifications, creating target equations, and exponent error analysis.
SECTION 3 KEY
Indiana Academic Standard 8.AF.2: Types of Solutions
Total: 4 Points
Problem 7: Classify 1 Point
\(5(2x - 4) = 10x - 12\)
1. Distribute left: \(10x - 20 = 10x - 12\)
2. Subtract \(10x\): \(-20 = -12\)
3. State: False statement (\(-20 \neq -12\))
4. Parallel lines; no value of \(x\) works.
Solution Type: No Solution (∅)
Problem 8: Classify 1 Point
\(3(4x + 2) - 6x = 2(3x + 3)\)
1. Left: \(12x + 6 - 6x = 6x + 6\)
2. Right: \(2(3x + 3) = 6x + 6\)
3. Compare: \(6x + 6 = 6x + 6 \implies 6 = 6\)
4. Identity: true for all values of \(x\).
Solution Type: Infinitely Many (∞)
Problem 9: Unique Solution 1 Point
\(2(3y - 1) = 4y + 10\)
1. Distribute: \(6y - 2 = 4y + 10\)
2. Subtract \(4y\): \(2y - 2 = 10\)
3. Add 2: \(2y = 12 \implies y = 6\)
Answers: One Solution; \(y = 6\)
Problem 10: Create Equations 1 Point
\(8x - 14 = \mathbf{A}x + \mathbf{B}\)
No Sol: \(\mathbf{A} = 8\), \(\mathbf{B} \neq -14\) (e.g., \(B = 0\) or \(5\))
Infinite: \(\mathbf{A} = 8\), \(\mathbf{B} = -14\) (exact identity)
Accept any valid constant \(B \neq -14\) for No Solution.
SECTION 4 KEY
Problem 11: Exponent Misconception Critique
Total: 2 Points
Conceptual Error Analysis:
Student's Mistake: The student subtracted positive \(2\) instead of negative \(2\). Under the quotient rule \(\frac{a^m}{a^n} = a^{m - n}\), subtracting a negative exponent requires \(3 - (-2) = 3 + 2 = 5\), not \(3 - 2 = 1\).
Teaching Tip: Show that \(\frac{1}{6^{-2}} = 6^2\). Therefore, \(\frac{6^3}{6^{-2}} = 6^3 \cdot 6^2 = 6^5\).
Correct Step-by-Step Derivation:
\(\frac{6^3}{6^{-2}} = 6^{3 - (-2)}\)
\(= 6^{3 + 2}\)
\(= 6^5\)
\(= 7,776\)
Correct Final Solution: \(6^5\) (or \(7,776\))
Intervention Action Guide: Scale: 8-10 Mastery • 6-7 Targeted Review • ≤5 Reteach Core
Students struggling with Section 1 need exponent expansion tables to see why \(a^{-n} = \frac{1}{a^n}\). Students struggling with Section 2 need two-color highlighters to separate variable terms from constant terms. Students struggling with Section 3 need the 3-scenario comparison chart (different slopes vs. same slope).
Algebra Foundations Solutions Key • Standard 8.AF.2 & Error Analysis Page 2 of 2
Algebra Triad Game Board Partner Practice Game • Grade 8 Mathematics
Algebra Triad: Four-in-a-Row Showdown
Game Board
Standards 8.NS.3 • 8.AF.1 • 8.AF.2
How to Play: Players take turns picking an open square. Both players solve on Page 2. If the active player solves correctly, they mark it (X or O ). If incorrect, their partner can steal. First to connect 4 in a row (horizontal, vertical, or diagonal) wins!
8.NS.3 Exponents
8.AF.1 Multi-Step Eq.
8.AF.2 Solution Types
A1 8.NS.3
\(3^4 \cdot 3^{-6}\)
Simplify power
A2 8.AF.1
\(2x + 5 = 4x - 7\)
Solve for \(x\)
A3 8.AF.2
\(3(x - 2) = 3x - 6\)
Classify: 1, 0, or ∞
A4 8.NS.3
\(\frac{7^{-2}}{7^{-5}}\)
Simplify power
A5 8.AF.1
\(3(y + 4) = 2y + 18\)
Solve for \(y\)
B1 8.AF.2
\(4x + 8 = 4x - 3\)
Classify: 1, 0, or ∞
B2 8.NS.3
\((2^3)^2 \cdot 2^{-4}\)
Evaluate value
B3 8.AF.1
\(-2(4m - 1) = -6m + 10\)
Solve for \(m\)
B4 8.AF.2
\(5x - 2 = 3x + 12\)
Classify & state \(x\)
B5 8.NS.3
\(\frac{10^2 \cdot 10^{-5}}{10^{-3}}\)
Evaluate value
C1 8.AF.1
\(5(w - 3) = 2w + 6\)
Solve for \(w\)
C2 8.AF.2
\(2(3x + 4) = 6x + 8\)
Classify: 1, 0, or ∞
C3 ☆ TRIAD WILD ☆
\(A \cdot B = 8\text{ for }8x + B = Ax - 2\)
Pick A & B for No Sol.
C4 8.NS.3
\((4^{-1})^{-3} \div 4^2\)
Evaluate value
C5 8.AF.1
\(4(2x - 1) = 3(2x + 2)\)
Solve for \(x\)
D1 8.NS.3
\(\frac{6^3 \cdot 6^{-1}}{6^4}\)
Fraction form
D2 8.AF.1
\(-3(x - 5) = 2x - 10\)
Solve for \(x\)
D3 8.AF.2
\(7x - 4 = 7x + 5\)
Classify: 1, 0, or ∞
D4 8.NS.3
\((5^2 \cdot 5^{-4})^0\)
Evaluate value
D5 8.AF.1
\(6 - 2k = 3k - 14\)
Solve for \(k\)
E1 8.AF.2
\(2(4x - 1) = 8x - 2\)
Classify: 1, 0, or ∞
E2 8.NS.3
\(\frac{3^{-3} \cdot 3^8}{3^2}\)
Single power
E3 8.AF.1
\(2(3x + 1) - 4 = 4x + 8\)
Solve for \(x\)
E4 8.AF.2
\(8x - 5 = 8x + 10\)
Classify: 1, 0, or ∞
E5 8.NS.3
\((2^{-2})^3 \cdot 2^9\)
Evaluate value
Player 1 (X):
Player 2 (O):
Winner:
Algebra Foundations Reteach • Standards 8.NS.3, 8.AF.1, & 8.AF.2 See Page 2 for Scratchwork Grid & Referee Answer Key →
Game Recording & Verification
Student Scratchwork & Referee Key
Page 2 of 2
Standards 8.NS.3 • 8.AF.1 • 8.AF.2
Player 1 Scratchwork (X) Show all steps
Square: _____
Square: _____
Square: _____
Square: _____
Player 2 Scratchwork (O) Show all steps
Square: _____
Square: _____
Square: _____
Square: _____
Referee Fast-Check Answer Key (Fold or Cover During Play)
All 25 Squares
Row A:
A1: \(3^{-2} = \frac{1}{9}\)
A2: \(x = 6\)
A3: Infinite (∞)
A4: \(7^3 = 343\)
A5: \(y = 6\)
Row B:
B1: No Sol. (∅)
B2: \(2^2 = 4\)
B3: \(m = -4\)
B4: One (\(x=7\))
B5: \(10^0 = 1\)
Row C:
C1: \(w = 7\)
C2: Infinite (∞)
C3: \(A=8, B\neq -2\)
C4: \(4^1 = 4\)
C5: \(x = 5\)
Row D:
D1: \(6^{-2} = \frac{1}{36}\)
D2: \(x = 5\)
D3: No Sol. (∅)
D4: \(5^0 = 1\)
D5: \(k = 4\)
Row E:
E1: Infinite (∞)
E2: \(3^3 = 27\)
E3: \(x = 5\)
E4: No Sol. (∅)
E5: \(2^3 = 8\)
Algebra Triad Game • Indiana Academic Standards 8.NS.3, 8.AF.1, & 8.AF.2 Post-Game Debrief: Identify which standard felt most challenging and review with Reteach Guide
Foundations Retest Practice Worksheet Indiana Academic Standards • Grade 8 Mathematics FORM B
Grade 8 Algebra Foundations
Page 1 of 2 (Front)
Standards 8.NS.3 & 8.AF.1
Student Name:
Date:
Period:
Self-Check:
/ 6 pts
SECTION 1
Indiana Academic Standard 8.NS.3
Properties of Exponents
Given a numeric expression with common rational number bases and integer exponents, apply the properties of exponents to generate equivalent expressions.
1. Product & Quotient Rules: Base 4
Simplify: \(\frac{4^5 \cdot 4^{-8}}{4^{-6}}\)
Single Power:
2. Power to a Power Rule: Mixed Bases
Simplify: \((3^2 \cdot 5^{-3})^2 \cdot 5^7\)
Simplified:
3. Scientific Application: Photons
An observatory telescope registers \((3^3)^4\) photon signals. It distributes them evenly across \(3^{14}\) sensor arrays: \(\frac{(3^3)^4}{3^{14}}\). Express the signals per array as a simplified fraction.
Final Value:
SECTION 2
Indiana Academic Standard 8.AF.1
Multi-Step Equations
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
4. Distribute & Collect: Solve for \(x\)
\(3(3x - 4) - 7 = 4x + 11\)
Solution:
5. Multi-Step with Negative: Solve for \(w\)
\(-4(2w + 3) = 3(w - 15)\)
Solution:
6. Streaming Plans: Real Context
StreamPlus charges \(\$15/\text{month} + \$30\) setup: \(15m + 30\). MovieMax charges \(\$20/\text{month}\) with no setup fee: \(20m\). Write and solve an equation to find the months, \(m\), when both cost the exact same.
Months \(m\):
Algebra Foundations Reteach • Form B • Standards 8.NS.3 & 8.AF.1 Turn over for Section 3 (8.AF.2) →
Indiana Academic Standards • Grade 8 Mathematics FORM B
Grade 8 Algebra Foundations
Page 2 of 2 (Back)
Standard 8.AF.2 & Error Analysis
Student Name: __________________________________________
Class Period: ___________
SECTION 3
Indiana Academic Standard 8.AF.2
Solution Classification
Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which is the case by successively transforming equations into simpler forms (\(x = a\), \(a = a\), or \(a = b\)).
7. Transform & Classify: Solve & State
\(4(3x - 2) = 12x - 15\)
Solution Type:
8. Transform & Classify: Solve & State
\(2(5x + 3) - 4x = 3(2x + 2)\)
Solution Type:
9. Solve & State Exact Value: Solve for \(y\)
\(3(2y - 3) = 2y + 11\)
Type:
Value of \(y\):
10. Create Target Equations: Find \(A\) & \(B\)
Given: \(6x - 18 = \mathbf{A}x + \mathbf{B}\) . Choose values of \(\mathbf{A}\) and \(\mathbf{B}\) for each condition below:
No Solution: \(A=\)___, \(B=\)___
Infinitely Many: \(A=\)___, \(B=\)___
Problem 11: Error Analysis & Exponent Misconception
Critique & Correct
Incorrect Student Claim:
A student evaluates: \((2^{-3})^2\) and writes:
"Power of power: \(2^{-3 \cdot 2} = 2^{-6}\)"
"Negative exponent makes it negative: \(2^{-6} = -64\)"
Explain the student's negative exponent mistake:
Your Correct Step-by-Step Work:
Apply \(a^{-n} = \frac{1}{a^n}\) and find the simplified positive fraction.
Correct Value:
Algebra Foundations Reteach • Form B • Standard 8.AF.2 Completed Practice Sheet • Check work with Form B Answer Key
Foundations Retest Solutions Key Teacher Resource • Grade 8 Worked Key FORM B KEY
Grade 8 Algebra Solutions Key
Key: Page 1 of 2
Standards 8.NS.3 & 8.AF.1
SECTION 1 KEY
Indiana Academic Standard 8.NS.3: Properties of Exponents
Total: 3 Points
Problem 1 Solution 1 Point
\(\frac{4^5 \cdot 4^{-8}}{4^{-6}}\)
1. Numerator product rule: \(4^{5 + (-8)} = 4^{-3}\)
2. Quotient rule: \(\frac{4^{-3}}{4^{-6}} = 4^{-3 - (-6)}\)
3. Subtracting negative: \(4^{-3 + 6} = 4^3\)
4. Evaluate: \(4^3 = 64\)
Correct Answer: \(4^3\) (or \(64\))
Error Note: Common error is \(-3 - (-6) \ne -9\). Result is \(+3\).
Problem 2 Solution 1 Point
\((3^2 \cdot 5^{-3})^2 \cdot 5^7\)
1. Power to a power: \(3^{2 \cdot 2} \cdot 5^{-3 \cdot 2} = 3^4 \cdot 5^{-6}\)
2. Multiply with \(5^7\): \(3^4 \cdot (5^{-6} \cdot 5^7)\)
3. Combine base 5: \(5^{-6 + 7} = 5^1 = 5\)
4. Evaluate: \(3^4 \cdot 5 = 81 \cdot 5 = 405\)
Correct Answer: \(3^4 \cdot 5\) or \(405\)
Error Note: Base 3 and base 5 must stay separated.
Problem 3 Solution (Telescope Photons) 1 Point
Expression: \(\frac{(3^3)^4}{3^{14}}\)
Step 1: \((3^3)^4 = 3^{3 \cdot 4} = 3^{12}\)
Step 2: \(\frac{3^{12}}{3^{14}} = 3^{12 - 14} = 3^{-2}\)
Step 3: Convert negative exponent: \(3^{-2} = \frac{1}{3^2} = \frac{1}{9}\)
Correct Answer: \(\frac{1}{9}\) (or \(3^{-2}\)) (signals per sensor)
SECTION 2 KEY
Indiana Academic Standard 8.AF.1: Multi-Step Linear Equations
Total: 3 Points
Problem 4 Solution 1 Point
\(3(3x - 4) - 7 = 4x + 11\)
1. Distribute: \(9x - 12 - 7 = 4x + 11\)
2. Combine left constants: \(9x - 19 = 4x + 11\)
3. Subtract \(4x\): \(5x - 19 = 11\)
4. Add 19: \(5x = 30 \implies x = 6\)
Correct Answer: \(x = 6\)
Problem 5 Solution 1 Point
\(-4(2w + 3) = 3(w - 15)\)
1. Distribute both: \(-8w - 12 = 3w - 45\)
2. Subtract \(3w\): \(-11w - 12 = -45\)
3. Add 12: \(-11w = -33\)
4. Divide by \(-11\): \(w = \frac{-33}{-11} = 3\)
Correct Answer: \(w = 3\)
Problem 6 Solution (Streaming Plans) 1 Point
Equation: \(15m + 30 = 20m\)
Subtract \(15m\): \(30 = 5m \implies m = 6\)
Check: StreamPlus \(= 15(6) + 30 = \$120\); MovieMax \(= 20(6) = \$120\) ✓
Correct Answer: \(m = 6\text{ months}\) (Total Cost = $120)
Algebra Foundations Solutions Key • Form B • Standards 8.NS.3 & 8.AF.1 Page 1 of 2
Teacher Resource • Grade 8 Worked Key FORM B KEY
Grade 8 Algebra Solutions Key
Key: Page 2 of 2
Standard 8.AF.2 & Error Analysis
Solutions for solution-type classifications, creating target equations, and negative exponent error critique.
SECTION 3 KEY
Indiana Academic Standard 8.AF.2: Types of Solutions
Total: 4 Points
Problem 7: Classify 1 Point
\(4(3x - 2) = 12x - 15\)
1. Distribute left: \(12x - 8 = 12x - 15\)
2. Subtract \(12x\): \(-8 = -15\)
3. State: False statement (\(-8 \neq -15\))
4. Parallel lines; contradiction.
Solution Type: No Solution (∅)
Problem 8: Classify 1 Point
\(2(5x + 3) - 4x = 3(2x + 2)\)
1. Left: \(10x + 6 - 4x = 6x + 6\)
2. Right: \(3(2x + 2) = 6x + 6\)
3. Compare: \(6x + 6 = 6x + 6 \implies 6 = 6\)
4. Identity: true for all real values of \(x\).
Solution Type: Infinitely Many (∞)
Problem 9: Unique Solution 1 Point
\(3(2y - 3) = 2y + 11\)
1. Distribute: \(6y - 9 = 2y + 11\)
2. Subtract \(2y\): \(4y - 9 = 11\)
3. Add 9: \(4y = 20 \implies y = 5\)
Answers: One Solution; \(y = 5\)
Problem 10: Create Equations 1 Point
\(6x - 18 = \mathbf{A}x + \mathbf{B}\)
No Sol: \(\mathbf{A} = 6\), \(\mathbf{B} \neq -18\) (e.g., \(B = 0\) or \(5\))
Infinite: \(\mathbf{A} = 6\), \(\mathbf{B} = -18\) (exact identity)
Accept any valid constant \(B \neq -18\) for No Solution.
SECTION 4 KEY
Problem 11: Exponent Misconception Critique
Total: 2 Points
Conceptual Error Analysis:
Student's Mistake: The student mistakenly assumed that a negative exponent produces a negative answer: \(2^{-6} \neq -64\). In reality, negative exponents represent repeated division / reciprocals: \(a^{-n} = \frac{1}{a^n}\).
Teaching Tip: Show the pattern: \(2^2 = 4\), \(2^1 = 2\), \(2^0 = 1\), \(2^{-1} = \frac{1}{2}\), \(2^{-2} = \frac{1}{4}\). The values are always positive fractions!
Correct Step-by-Step Derivation:
\((2^{-3})^2 = 2^{-3 \cdot 2}\)
\(= 2^{-6}\)
\(= \frac{1}{2^6}\)
\(= \frac{1}{64}\)
Correct Final Solution: \(\frac{1}{64}\) (or \(2^{-6}\))
Intervention Action Guide: Scale: 8-10 Mastery • 6-7 Targeted Review • ≤5 Reteach Core
Compare Form A vs. Form B scores. Students demonstrating growth from ≤5 to ≥8 have solidified mastery. Students continuing to struggle with Section 3 should pair up on the Algebra Triad Game Board.
Algebra Foundations Solutions Key • Form B • Standard 8.AF.2 & Error Analysis Page 2 of 2
Algebra Triad Encore Game Board Partner Practice Game • Grade 8 Mathematics FORM B
Algebra Triad: Four-in-a-Row Showdown
Game Board (Form B)
Standards 8.NS.3 • 8.AF.1 • 8.AF.2
How to Play: Players take turns picking an open square. Both players solve on Page 2. If the active player solves correctly, they mark it (X or O ). If incorrect, their partner can steal. First to connect 4 in a row (horizontal, vertical, or diagonal) wins!
8.NS.3 Exponents
8.AF.1 Multi-Step Eq.
8.AF.2 Solution Types
A1 8.NS.3
\(4^6 \cdot 4^{-8}\)
Simplify power
A2 8.AF.1
\(3x + 8 = 5x - 4\)
Solve for \(x\)
A3 8.AF.2
\(4(2x - 1) = 8x - 4\)
Classify: 1, 0, or ∞
A4 8.NS.3
\(\frac{5^{-3}}{5^{-6}}\)
Simplify power
A5 8.AF.1
\(2(y + 5) = 3y + 2\)
Solve for \(y\)
B1 8.AF.2
\(6x + 9 = 6x - 2\)
Classify: 1, 0, or ∞
B2 8.NS.3
\((3^2)^3 \cdot 3^{-5}\)
Evaluate value
B3 8.AF.1
\(-3(2m - 4) = -4m + 16\)
Solve for \(m\)
B4 8.AF.2
\(4x - 6 = 2x + 10\)
Classify & state \(x\)
B5 8.NS.3
\(\frac{2^4 \cdot 2^{-7}}{2^{-3}}\)
Evaluate value
C1 8.AF.1
\(4(w - 2) = 2w + 10\)
Solve for \(w\)
C2 8.AF.2
\(5(2x + 3) = 10x + 15\)
Classify: 1, 0, or ∞
C3 ☆ TRIAD WILD ☆
\(10x + B = Ax - 4\)
Pick A & B for No Sol.
C4 8.NS.3
\((2^{-2})^{-2} \div 2^3\)
Evaluate value
C5 8.AF.1
\(3(3x - 2) = 2(4x + 1)\)
Solve for \(x\)
D1 8.NS.3
\(\frac{8^4 \cdot 8^{-2}}{8^3}\)
Fraction form
D2 8.AF.1
\(-2(x - 6) = 3x - 13\)
Solve for \(x\)
D3 8.AF.2
\(9x - 7 = 9x + 2\)
Classify: 1, 0, or ∞
D4 8.NS.3
\((6^3 \cdot 6^{-5})^0\)
Evaluate value
D5 8.AF.1
\(12 - 3k = 2k - 18\)
Solve for \(k\)
E1 8.AF.2
\(3(3x - 2) = 9x - 6\)
Classify: 1, 0, or ∞
E2 8.NS.3
\(\frac{4^{-2} \cdot 4^7}{4^3}\)
Single power
E3 8.AF.1
\(3(2x + 4) - 5 = 5x + 11\)
Solve for \(x\)
E4 8.AF.2
\(5x - 3 = 5x + 8\)
Classify: 1, 0, or ∞
E5 8.NS.3
\((3^{-1})^4 \cdot 3^6\)
Evaluate value
Player 1 (X):
Player 2 (O):
Winner:
Algebra Triad Game • Form B • Standards 8.NS.3, 8.AF.1, & 8.AF.2 See Page 2 for Scratchwork Grid & Referee Answer Key →
Game Recording & Verification FORM B
Student Scratchwork & Referee Key
Page 2 of 2
Standards 8.NS.3 • 8.AF.1 • 8.AF.2
Player 1 Scratchwork (X) Show all steps
Square: _____
Square: _____
Square: _____
Square: _____
Player 2 Scratchwork (O) Show all steps
Square: _____
Square: _____
Square: _____
Square: _____
Form B Referee Fast-Check Key (Fold or Cover During Play)
All 25 Squares
Row A:
A1: \(4^{-2} = \frac{1}{16}\)
A2: \(x = 6\)
A3: Infinite (∞)
A4: \(5^3 = 125\)
A5: \(y = 8\)
Row B:
B1: No Sol. (∅)
B2: \(3^1 = 3\)
B3: \(m = -2\)
B4: One (\(x=8\))
B5: \(2^0 = 1\)
Row C:
C1: \(w = 9\)
C2: Infinite (∞)
C3: \(A=10, B\neq -4\)
C4: \(2^1 = 2\)
C5: \(x = 8\)
Row D:
D1: \(8^{-1} = \frac{1}{8}\)
D2: \(x = 5\)
D3: No Sol. (∅)
D4: \(6^0 = 1\)
D5: \(k = 6\)
Row E:
E1: Infinite (∞)
E2: \(4^2 = 16\)
E3: \(x = 4\)
E4: No Sol. (∅)
E5: \(3^2 = 9\)
Algebra Triad Game • Form B • Indiana Standards 8.NS.3, 8.AF.1, & 8.AF.2 Post-Game Debrief: Identify which standard felt most challenging and review with Reteach Guide
