Unit 8 · Teacher Answer Key
Unit 8 Projects — Teacher Answer Key
Equations & Inequalities · Grade 6 · Standards 6.EE.5, 6.EE.6, 6.EE.7, 6.EE.8
For teachers only — do not distribute to students.
Worked solutions below use each project's default input values.
Students who change the inputs will get different but mathematically similar answers;
the same reasoning steps apply. Use the "Sample Expert Response" paragraphs as
4-point anchor papers.
Phase 1 — Translate the Clue & Build Lock B 6.EE.6 · Writing Equations
Lock A — Story Clue (hard-coded):
x + 7 = 20
Clue: "When you add 7 to the combination, you get 20."
Variable: x = the unknown combination number.
Equation: x + 7 = 20
(Also accepted: 7 + x = 20)
Variable: x = the unknown combination number.
Equation: x + 7 = 20
(Also accepted: 7 + x = 20)
Lock A equation: x + 7 = 20
Lock B — Customizable Clue (default p = 9, q = 25):
p (added) = 9
q (result) = 25
x + 9 = 25
Clue built by JS: "When you add 9 to the combination, you get 25."
Equation: x + 9 = 25
Solve (inverse operation — subtract p from both sides):
x = q − p = 25 − 9 = 16
Substitution check: 16 + 9 = 25 ✓ Verified
Equation: x + 9 = 25
Solve (inverse operation — subtract p from both sides):
x = q − p = 25 − 9 = 16
Substitution check: 16 + 9 = 25 ✓ Verified
Lock B combination: x = 16
Phase 2 — Solve Lock C (Addition Equation) 6.EE.7 · Add/Subtract Equations
Default inputs:
p (added to x) = 8
q (right side) = 15
x + 8 = 15
Equation: x + 8 = 15
Inverse operation: Addition is undone by subtraction.
Subtract 8 from both sides:
x + 8 − 8 = 15 − 8
x = 7
Substitution check (verification):
Substitute x = 7 back into the original equation:
7 + 8 = 15 ✓ True — combination is correct
Inverse operation: Addition is undone by subtraction.
Subtract 8 from both sides:
x + 8 − 8 = 15 − 8
x = 7
Substitution check (verification):
Substitute x = 7 back into the original equation:
7 + 8 = 15 ✓ True — combination is correct
Lock C combination: x = 7
Phase 3 — Solve Lock D (Multiplication Equation) 6.EE.7 · Multiply/Divide Equations
Default inputs:
p (multiplier) = 4
q (right side) = 28
4x = 28
Equation: 4x = 28
Inverse operation: Multiplication is undone by division.
Divide both sides by 4:
4x ÷ 4 = 28 ÷ 4
x = 7
Divisibility check: 28 ÷ 4 = 7 exactly (no remainder) — the solution is a whole number.
Substitution check (verification):
Substitute x = 7: 4 × 7 = 28 ✓ True — combination is correct
Inverse operation: Multiplication is undone by division.
Divide both sides by 4:
4x ÷ 4 = 28 ÷ 4
x = 7
Divisibility check: 28 ÷ 4 = 7 exactly (no remainder) — the solution is a whole number.
Substitution check (verification):
Substitute x = 7: 4 × 7 = 28 ✓ True — combination is correct
Lock D combination: x = 7
Phase 4 — Sequence the Puzzles & Final Quick-Check 6.EE.7 · Verify
Suggested puzzle sequence (from project placeholder):
Lock A (x + 7 = 20) first — simplest translation exercise, introduces the concept of a variable.
Lock B (x + 9 = 25) second — similar structure, student-generated numbers, builds confidence.
Lock C (x + 8 = 15) third — same operation type but solver must execute the inverse step independently.
Lock D (4x = 28) last — multiplication equation introduces a new inverse operation (division); highest cognitive demand.
Lock B (x + 9 = 25) second — similar structure, student-generated numbers, builds confidence.
Lock C (x + 8 = 15) third — same operation type but solver must execute the inverse step independently.
Lock D (4x = 28) last — multiplication equation introduces a new inverse operation (division); highest cognitive demand.
Final Quick-Check: Solve 3x = 21.
Equation: 3x = 21
Divide both sides by 3:
x = 21 ÷ 3 = 7
Substitution check: 3 × 7 = 21 ✓
Divide both sides by 3:
x = 21 ÷ 3 = 7
Substitution check: 3 × 7 = 21 ✓
Correct answer: x = 7
JavaScript check: if (v === 7)
Complete Answer Key Summary — All Four Locks (Default Values)
Lock A: x + 7 = 20 → x = 20 − 7 = 13 (verify: 13 + 7 = 20 ✓)
Lock B: x + 9 = 25 → x = 25 − 9 = 16 (verify: 16 + 9 = 25 ✓)
Lock C: x + 8 = 15 → x = 15 − 8 = 7 (verify: 7 + 8 = 15 ✓)
Lock D: 4x = 28 → x = 28 ÷ 4 = 7 (verify: 4 × 7 = 28 ✓)
Lock B: x + 9 = 25 → x = 25 − 9 = 16 (verify: 16 + 9 = 25 ✓)
Lock C: x + 8 = 15 → x = 15 − 8 = 7 (verify: 7 + 8 = 15 ✓)
Lock D: 4x = 28 → x = 28 ÷ 4 = 7 (verify: 4 × 7 = 28 ✓)
Note: Lock A's combination (13) differs from Lock B's (16). Lock C and Lock D both happen to give x = 7 with the default numbers — this is coincidental and will differ when students use their own inputs.
Sample Expert Response (4/4 Deliverable)
Sample Escape Room Blueprint using default values:
"My escape room has four locks. Lock A equation: x + 7 = 20; solve by subtracting 7 from both sides: x = 13; verify: 13 + 7 = 20. Lock B equation: x + 9 = 25; solve by subtracting 9: x = 16; verify: 16 + 9 = 25. Lock C equation: x + 8 = 15; solve by subtracting 8: x = 7; verify: 7 + 8 = 15. Lock D equation: 4x = 28; solve by dividing both sides by 4: x = 7; verify: 4 × 7 = 28. I ordered the locks by difficulty — addition equations first since they all use the same inverse operation, then multiplication last because it introduces division as the inverse."
"My escape room has four locks. Lock A equation: x + 7 = 20; solve by subtracting 7 from both sides: x = 13; verify: 13 + 7 = 20. Lock B equation: x + 9 = 25; solve by subtracting 9: x = 16; verify: 16 + 9 = 25. Lock C equation: x + 8 = 15; solve by subtracting 8: x = 7; verify: 7 + 8 = 15. Lock D equation: 4x = 28; solve by dividing both sides by 4: x = 7; verify: 4 × 7 = 28. I ordered the locks by difficulty — addition equations first since they all use the same inverse operation, then multiplication last because it introduces division as the inverse."
Sample 4 / 3 / 2 Rubric Scoring Guide:
4 — Expert: All four lock equations written correctly using a variable; each solution found by explicitly naming the inverse operation used; substitution check shown for every lock; sequence reasoning addresses difficulty level.
3 — Proficient: All four solutions correct; at least one substitution check shown; blueprint lists equations and solutions for every lock.
2 — Developing: One or two equations written correctly; solutions attempted but one has a computation error; blueprint is missing one or more locks' equations or solutions.
4 — Expert: All four lock equations written correctly using a variable; each solution found by explicitly naming the inverse operation used; substitution check shown for every lock; sequence reasoning addresses difficulty level.
3 — Proficient: All four solutions correct; at least one substitution check shown; blueprint lists equations and solutions for every lock.
2 — Developing: One or two equations written correctly; solutions attempted but one has a computation error; blueprint is missing one or more locks' equations or solutions.
Phase 1 — How Much More Do We Need? 6.EE.7 · Addition Equation
Default inputs:
Amount already raised = $85
Fundraiser goal = $300
$85 + x = $300
Equation model: raised + x = goal
85 + x = 300
Inverse operation: Subtraction undoes addition. Subtract 85 from both sides:
x = 300 − 85 = $215
Substitution check:
$85 + $215 = $300 ✓ Verified
Interpretation: The class still needs to raise $215 to meet the $300 goal.
85 + x = 300
Inverse operation: Subtraction undoes addition. Subtract 85 from both sides:
x = 300 − 85 = $215
Substitution check:
$85 + $215 = $300 ✓ Verified
Interpretation: The class still needs to raise $215 to meet the $300 goal.
x = $215 more needed
Phase 2 — Plan the Sales 6.EE.7 · Multiplication Equation
Default inputs:
Price per item = $5
Sales target = $150
n × $5 = $150
Equation model: n × price = target (n = number of items to sell)
5n = 150
Inverse operation: Division undoes multiplication. Divide both sides by 5:
n = 150 ÷ 5 = 30 items
Divisibility check: 150 ÷ 5 = 30 exactly (no remainder) — whole number answer. ✓
Substitution check:
30 × $5 = $150 ✓ Verified
Interpretation: The class must sell 30 items at $5 each to earn $150 from the bake sale.
5n = 150
Inverse operation: Division undoes multiplication. Divide both sides by 5:
n = 150 ÷ 5 = 30 items
Divisibility check: 150 ÷ 5 = 30 exactly (no remainder) — whole number answer. ✓
Substitution check:
30 × $5 = $150 ✓ Verified
Interpretation: The class must sell 30 items at $5 each to earn $150 from the bake sale.
n = 30 items to sell
Phase 3 — Set the Goal Condition (Inequality) 6.EE.8 · Writing & Graphing Inequalities
Default inputs:
Minimum goal = $300
Symbol = ≥ (greater than or equal to)
x ≥ 300
Inequality written: x ≥ 300
Read as: "The total amount raised (x) must be at least $300."
Solution set: All values of x that are 300 or greater.
x ∈ {300, 301, 302, 305, 310, 400, 500, …} — infinitely many solutions.
Number-line description: Draw a closed circle at 300 (closed = includes 300, because ≥ means "equal to or greater than") and shade the ray extending to the right (toward larger numbers).
If the symbol were strictly >, use an open circle at 300.
Sample value check (from JS logic):
x = 298: 298 ≥ 300 → FALSE (does not satisfy — below goal)
x = 300: 300 ≥ 300 → TRUE (satisfies — exactly meets goal)
x = 302: 302 ≥ 300 → TRUE (satisfies — exceeds goal)
x = 304: 304 ≥ 300 → TRUE (satisfies)
Read as: "The total amount raised (x) must be at least $300."
Solution set: All values of x that are 300 or greater.
x ∈ {300, 301, 302, 305, 310, 400, 500, …} — infinitely many solutions.
Number-line description: Draw a closed circle at 300 (closed = includes 300, because ≥ means "equal to or greater than") and shade the ray extending to the right (toward larger numbers).
If the symbol were strictly >, use an open circle at 300.
Sample value check (from JS logic):
x = 298: 298 ≥ 300 → FALSE (does not satisfy — below goal)
x = 300: 300 ≥ 300 → TRUE (satisfies — exactly meets goal)
x = 302: 302 ≥ 300 → TRUE (satisfies — exceeds goal)
x = 304: 304 ≥ 300 → TRUE (satisfies)
Inequality: x ≥ 300 | Closed circle at 300, shade right
Phase 4 — Will the Plan Hit the Goal? & Quick Checks 6.EE.7 · 6.EE.8 · Decision & Verify
Decision using default values:
Current raised: $85
Sales plan: 30 items × $5 = $150
Projected total: $85 + $150 = $235
Goal condition: x ≥ $300
Check: $235 ≥ $300? → FALSE (235 < 300)
Conclusion with default values: The plan will not meet the goal. The class would raise $235 total, which is $65 short. They need to sell more items or find additional funding.
Sales plan: 30 items × $5 = $150
Projected total: $85 + $150 = $235
Goal condition: x ≥ $300
Check: $235 ≥ $300? → FALSE (235 < 300)
Conclusion with default values: The plan will not meet the goal. The class would raise $235 total, which is $65 short. They need to sell more items or find additional funding.
Final Quick-Check: Solve 5x = 60.
Equation: 5x = 60
Divide both sides by 5:
x = 60 ÷ 5 = 12
Substitution check: 5 × 12 = 60 ✓
Divide both sides by 5:
x = 60 ÷ 5 = 12
Substitution check: 5 × 12 = 60 ✓
Correct answer: x = 12
JavaScript check: if (v === 12)
Inequality Quick-Check: Does x = 8 satisfy x ≥ 8? Does x = 5 satisfy x ≥ 8? Explain why or why not.
x = 8: Check 8 ≥ 8 → TRUE (8 equals 8, and ≥ includes "equal to"). Satisfies.
x = 5: Check 5 ≥ 8 → FALSE (5 is less than 8). Does not satisfy.
Model answer: "x = 8 satisfies x ≥ 8 because 8 ≥ 8 is true — 8 is not less than itself. x = 5 does not satisfy x ≥ 8 because 5 is less than 8."
x = 5: Check 5 ≥ 8 → FALSE (5 is less than 8). Does not satisfy.
Model answer: "x = 8 satisfies x ≥ 8 because 8 ≥ 8 is true — 8 is not less than itself. x = 5 does not satisfy x ≥ 8 because 5 is less than 8."
x = 8: satisfies | x = 5: does not satisfy
JS check looks for: mentions of "8" and "5", words indicating truth ("satisf"/"true"/"yes"), and words indicating falsity ("not"/"false"/"no"/"less").
Sample Expert Response (4/4 Deliverable)
Sample Fundraiser Plan using default values:
"Our fundraiser goal is $300. We have raised $85, so the equation 85 + x = 300 gives x = $215 — we still need $215 more. To reach a sales target of $150, we must sell 30 items at $5 each (equation: 5n = 150, verified: 30 × 5 = 150). The condition for success is x ≥ $300, meaning total raised must be at least $300; on a number line this is a closed circle at 300 shaded to the right. Based on our plan, we will not meet the goal because $85 + $150 = $235, and 235 < 300 — we need to sell approximately 13 more items or find an additional $65 to close the gap."
"Our fundraiser goal is $300. We have raised $85, so the equation 85 + x = 300 gives x = $215 — we still need $215 more. To reach a sales target of $150, we must sell 30 items at $5 each (equation: 5n = 150, verified: 30 × 5 = 150). The condition for success is x ≥ $300, meaning total raised must be at least $300; on a number line this is a closed circle at 300 shaded to the right. Based on our plan, we will not meet the goal because $85 + $150 = $235, and 235 < 300 — we need to sell approximately 13 more items or find an additional $65 to close the gap."
Sample 4 / 3 / 2 Rubric Scoring Guide:
4 — Expert: Addition equation written, solved, and verified with substitution; multiplication equation written, solved, and verified; inequality written with correct symbol, solution set described in words, number-line circle type (open/closed) justified; conclusion explicitly compares projected total to goal.
3 — Proficient: All three equations/inequality solved correctly; at least one substitution check shown; plan reaches a conclusion using real numbers.
2 — Developing: One or two equations solved correctly; inequality symbol correct but solution set or number-line description incomplete; conclusion present but may not use actual numbers.
4 — Expert: Addition equation written, solved, and verified with substitution; multiplication equation written, solved, and verified; inequality written with correct symbol, solution set described in words, number-line circle type (open/closed) justified; conclusion explicitly compares projected total to goal.
3 — Proficient: All three equations/inequality solved correctly; at least one substitution check shown; plan reaches a conclusion using real numbers.
2 — Developing: One or two equations solved correctly; inequality symbol correct but solution set or number-line description incomplete; conclusion present but may not use actual numbers.