Algebra Quiz: Modeling With Equation Inequalityconstraints

What this quiz covers

This quiz focuses on Modeling With Equation Inequalityconstraints, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Viable; it satisfies $x+y\le 30$, $x\ge 10$, $2x+3y\le 75$, and uses nonnegative whole numbers. (correct answer)
2. Nonviable; it violates the space constraint because $12+6>30$.
3. Nonviable; it violates the watering constraint because $2(12)+3(6)>75$.
4. Nonviable; it violates the minimum tomato constraint because $12<10$.

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. A solution is viable (feasible) if it satisfies EVERY SINGLE constraint AND makes sense in the real-world context (like no negative quantities, whole items when you can't buy half an item, etc.). If even one constraint is violated, or if the solution is unrealistic, it's nonviable. Think of constraints like security checkpoints—you need to pass through all of them! To check if (12, 6) is viable, we substitute into each constraint: Space constraint: 12 + 6 = 18 ≤ 30 ✓ (true). Minimum tomatoes: 12 ≥ 10 ✓ (true). Watering time: 2(12) + 3(6) = 24 + 18 = 42 ≤ 75 ✓ (true). Also checking context: 12 and 6 are both nonnegative whole numbers ✓. Conclusion: viable because all constraints are satisfied. Choice A is correct because it properly checks all constraints showing (12, 6) is viable and correctly identifies that all inequalities are satisfied. Choice B says the point is nonviable but doesn't check correctly: it claims 12 + 6 > 30, but 12 + 6 = 18, which is less than 30. To determine viability, you MUST check every single constraint accurately—missing even one or calculating incorrectly can lead to rejecting a perfectly good solution! The viability-checking procedure: Make a checklist of every constraint. For each one, substitute the point and check if it's satisfied. Write 'Yes' or 'No' next to each constraint. If even one 'No' appears, the solution is nonviable—identify which constraint(s) failed. Also do a reality check: negative quantities? fractional items when must be whole? These context violations also make solutions nonviable!

Question 2

1. $\{6x+4y\ge 60,\ x+y\ge 12,\ x\ge 0,\ y\ge 0\}$
2. $\{6x+4y\le 60,\ x+y\le 12,\ x\ge 0,\ y\ge 0\}$
3. $\{6x+4y\le 60,\ x+y\ge 12,\ x\ge 0,\ y\ge 0\}$ (correct answer)
4. $\{10x+4y\le 60,\ x+y\ge 12,\ x\ge 0,\ y\ge 0\}$

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. When setting up constraints from context, look for key phrases: 'at most' or 'no more than' → ≤, 'at least' or 'minimum' → ≥, 'exactly' or 'must be' → =, 'less than' → <, 'more than' → >. Also, don't forget implicit constraints like x ≥ 0 and y ≥ 0 (can't have negative quantities) or x, y must be integers (if counting discrete items). Let's identify all the constraints from the snack-buying scenario: the budget 'at most $60' gives us 6x + 4y ≤ 60, the requirement 'at least 12 total boxes' gives us x + y ≥ 12, and non-negativity gives us x ≥ 0, y ≥ 0. Putting this together as a system: {6x + 4y ≤ 60, x + y ≥ 12, x ≥ 0, y ≥ 0}. This system captures all the limitations and requirements of the situation. Choice C is correct because it includes all constraints with correct inequality directions. Choice B is tempting but has the total boxes inequality backwards: 'at least 12' means ≥ 12, not ≤ 12, which would incorrectly limit to no more than 12 boxes instead of requiring a minimum. The constraint-writing recipe: (1) List EVERY limitation mentioned in the problem (budget, time, capacity, minimums, etc.), (2) For each, identify the inequality symbol from key words ('at most' → ≤, 'at least' → ≥, etc.), (3) Write the mathematical inequality using the costs, rates, or quantities from context, (4) Don't forget implicit constraints like x ≥ 0, y ≥ 0 (non-negativity) or x, y integers (if discrete). Make sure your system captures EVERY constraint!

Question 3

1. $(150,120)$
2. $(100,200)$
3. $(210,110)$ (correct answer)
4. $(0,120)$

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. A solution is viable (feasible) if it satisfies EVERY SINGLE constraint AND makes sense in the real-world context (like no negative quantities, whole items when you can't buy half an item, etc.). If even one constraint is violated, or if the solution is unrealistic, it's nonviable. Think of constraints like security checkpoints—you need to pass through all of them! Let's check each point against all constraints: For (150, 120): Total tickets: 150 + 120 = 270 ≤ 300 ✓, Floor minimum: 120 ≥ 120 ✓, Non-negative: both ≥ 0 ✓ → viable. For (100, 200): Total tickets: 100 + 200 = 300 ≤ 300 ✓, Floor minimum: 200 ≥ 120 ✓, Non-negative: both ≥ 0 ✓ → viable. For (210, 110): Total tickets: 210 + 110 = 320 ≤ 300 ✗ (320 > 300), Floor minimum: 110 ≥ 120 ✗ (110 < 120) → nonviable (violates both!). For (0, 120): Total tickets: 0 + 120 = 120 ≤ 300 ✓, Floor minimum: 120 ≥ 120 ✓, Non-negative: both ≥ 0 ✓ → viable. Notice how (210, 110) fails because it violates BOTH constraints—too many total tickets AND not enough floor tickets! Choice C is correct because it correctly identifies the point (210, 110) as nonviable due to violating the total tickets constraint (320 > 300) and the floor tickets minimum (110 < 120). Choice A would be viable but (150, 120) actually satisfies all constraints: 150 + 120 = 270 ≤ 300 and 120 ≥ 120. Don't be fooled—a point needs to violate at least one constraint to be nonviable! The viability-checking procedure: Make a checklist of every constraint. For each one, substitute the point and check if it's satisfied. Write 'Yes' or 'No' next to each constraint. If even one 'No' appears, the solution is nonviable—identify which constraint(s) failed. When checking viability, substitute carefully: if the point is (210, 110), that means x = 210 and y = 110. Substitute those values into EVERY inequality and equation. It's tedious but necessary—one missed check could mean accepting an infeasible solution!

Question 4

1. Nonviable; it violates the whole-number requirement even though it satisfies the inequalities. (correct answer)
2. Nonviable; it violates $x+y\ge 8$ because $3.5+6=7.5<8$.
3. Viable; it satisfies $4x+y\le 25$ and $x+y\ge 8$.
4. Nonviable; it violates the budget constraint because $4(3.5)+6=26>25$.

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. A solution is viable (feasible) if it satisfies EVERY SINGLE constraint AND makes sense in the real-world context (like no negative quantities, whole items when you can't buy half an item, etc.). If even one constraint is violated, or if the solution is unrealistic, it's nonviable. Think of constraints like security checkpoints—you need to pass through all of them! To check if (3.5, 6) is viable, we substitute into each constraint: for 4x + y ≤ 25, 4(3.5) + 6 = 14 + 6 = 20 ≤ 25 (true); for x + y ≥ 8, 3.5 + 6 = 9.5 ≥ 8 (true); for x ≥ 0, y ≥ 0 (true). But checking context: must be whole numbers, and 3.5 is fractional, which is impossible for items. Conclusion: nonviable because violates whole-number requirement. Choice C is correct because it correctly identifies that it violates the whole-number requirement even though it satisfies the inequalities. Choice B says it's nonviable but incorrectly checks the budget: 20 ≤ 25 is satisfied, not violated. Remember, mathematical satisfaction isn't enough—context matters! The viability-checking procedure: Make a checklist of every constraint. For each one, substitute the point and check if it's satisfied. Write 'Yes' or 'No' next to each constraint. If even one 'No' appears, the solution is nonviable—identify which constraint(s) failed. Also do a reality check: negative quantities? fractional items when must be whole? These context violations also make solutions nonviable! Remember: viable means 'it could actually happen in real life.' Mathematically satisfying inequalities is necessary, but not sufficient. Ask yourself: In this context, can quantities be negative? Can they be fractions? Are there other real-world restrictions? A solution can satisfy all the math but still be nonviable if it violates reality!

Question 5

1. $\{3x+2y\le 60,\; x+y\le 25,\; x\ge 0,\; y\ge 0\}$
2. $\{3x+2y\ge 60,\; x+y\ge 25,\; x\ge 0,\; y\ge 0\}$
3. $\{3x+2y\le 60,\; x+y\ge 25,\; x\ge 0,\; y\ge 0\}$ (correct answer)
4. $\{3x+2y\le 60,\; x+y\ge 25\}$

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. When setting up constraints from context, look for key phrases: 'at most' or 'no more than' → ≤, 'at least' or 'minimum' → ≥, 'exactly' or 'must be' → =, 'less than' → <, 'more than' → >. Also, don't forget implicit constraints like x ≥ 0 and y ≥ 0 (can't have negative quantities) or x, y must be integers (if counting discrete items). Let's identify all the constraints from the fundraiser situation: the budget 'at most $60' gives us 3x + 2y ≤ 60, the requirement 'at least 25 total items' gives us x + y ≥ 25, and the implicit non-negativity gives us x ≥ 0 and y ≥ 0. Putting this together as a system: {3x + 2y ≤ 60, x + y ≥ 25, x ≥ 0, y ≥ 0}. This system captures all the limitations and requirements of the situation. Choice C is correct because it includes all constraints with correct inequality directions. Choice A has the inequality direction backwards: 'at least 25' means ≥, not ≤. This is a super common error! When total items 'at least 25,' we need x + y ≥ 25, meaning we can't fall below 25. The 'at least' puts a floor, not a ceiling. The constraint-writing recipe: (1) List EVERY limitation mentioned in the problem (budget, time, capacity, minimums, etc.), (2) For each, identify the inequality symbol from key words ('at most' → ≤, 'at least' → ≥, etc.), (3) Write the mathematical inequality using the costs, rates, or quantities from context, (4) Don't forget implicit constraints like x ≥ 0, y ≥ 0 (non-negativity) or x, y integers (if discrete). Make sure your system captures EVERY constraint!

Question 6

1. It violates no constraints (it is viable).
2. It violates both $3x+y\le 120$ and $x+y\ge 20$.
3. It violates $y\le 50$ only.
4. It violates $3x+y\le 120$ only. (correct answer)

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. Checking viability is systematic: (1) Substitute the proposed solution into each constraint, (2) Verify each inequality or equation is satisfied, (3) Check context realism (non-negative? whole numbers if needed?). If everything checks out, it's viable. If anything fails, it's nonviable—and you should explain which constraint was violated. To check if (30, 40) is viable, we substitute into each constraint: Checking 3x + y ≤ 120: 3(30) + 40 = 90 + 40 = 130 ≤ 120 ✗ (false, since 130 > 120). Checking x + y ≥ 20: 30 + 40 = 70 ≥ 20 ✓ (true). Checking y ≤ 50: 40 ≤ 50 ✓ (true). Checking x ≥ 0: 30 ≥ 0 ✓ and y ≥ 0: 40 ≥ 0 ✓. Conclusion: nonviable because it violates the time constraint 3x + y ≤ 120. Choice A is correct because it properly checks all constraints showing that only 3x + y ≤ 120 is violated (130 > 120). Choice D says the point is viable but doesn't check all constraints: it misses that 3(30) + 40 = 130 exceeds the 120-minute limit. To determine viability, you MUST check every single constraint—missing even one can lead to accepting an impossible solution! When checking viability, substitute carefully: if the point is (30, 40), that means x = 30 and y = 40. Substitute those values into EVERY inequality and equation. For example, if one constraint is 3x + y ≤ 120, check: 3(30) + 40 = 90 + 40 = 130 ≤ 120? No! Do this for every single constraint. It's tedious but necessary—one missed check could mean accepting an infeasible solution!

Question 7

1. $\{3x+2y\ge 60,\ x+y\ge 10,\ x\ge 0,\ y\ge 0\}$
2. $\{3x+2y\le 60,\ x+y\ge 10,\ x\ge 0,\ y\ge 0\}$ (correct answer)
3. $\{3x+2y\le 60,\ x+y\le 10,\ x\ge 0,\ y\ge 0\}$
4. $\{5x+2y\le 60,\ x+y\ge 10,\ x\ge 0,\ y\ge 0\}$

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. When setting up constraints from context, look for key phrases: 'at most' or 'no more than' → ≤, 'at least' or 'minimum' → ≥, 'exactly' or 'must be' → =, 'less than' → <, 'more than' → >. Also, don't forget implicit constraints like x ≥ 0 and y ≥ 0 (can't have negative quantities) or x, y must be integers (if counting discrete items). Let's identify all the constraints from 'notebooks cost $3 each, pens cost$2 each, at most $60 to spend, need at least 10 total items': The budget constraint 'at most$60' gives us 3x + 2y ≤ 60. The minimum items constraint 'at least 10 total items' gives us x + y ≥ 10. The non-negativity constraints (can't buy negative items) give us x ≥ 0 and y ≥ 0. Putting this together as a system: {3x + 2y ≤ 60, x + y ≥ 10, x ≥ 0, y ≥ 0}. This system captures all the limitations and requirements of the situation. Choice B is correct because it includes all constraints with correct inequality directions: the budget constraint uses ≤ (at most), the minimum items constraint uses ≥ (at least), and non-negativity constraints are included. Choice A has the inequality direction backwards: 'at most $60' means ≤ (can equal or be less), not ≥. This is a super common error! When total cost 'at most$60,' we need cost ≤ 60, meaning we can't exceed 60. The 'at most' puts a ceiling, not a floor. The constraint-writing recipe: (1) List EVERY limitation mentioned in the problem (budget, time, capacity, minimums, etc.), (2) For each, identify the inequality symbol from key words ('at most' → ≤, 'at least' → ≥, etc.), (3) Write the mathematical inequality using the costs, rates, or quantities from context, (4) Don't forget implicit constraints like x ≥ 0, y ≥ 0 (non-negativity) or x, y integers (if discrete). Make sure your system captures EVERY constraint!

Question 8

1. Viable; it satisfies all constraints. (correct answer)
2. Nonviable; it violates $2x+y\le 30$ because $2(9)+9=27$.
3. Nonviable; it violates $x+y\ge 18$ because $9+9=16$.
4. Nonviable; it violates $x\le 10$ because $9>10$.

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. A solution is viable (feasible) if it satisfies EVERY SINGLE constraint AND makes sense in the real-world context (like no negative quantities, whole items when you can't buy half an item, etc.). If even one constraint is violated, or if the solution is unrealistic, it's nonviable. Think of constraints like security checkpoints—you need to pass through all of them! To check if (9, 9) is viable, we substitute into each constraint: Checking 2x + y ≤ 30: 2(9) + 9 = 18 + 9 = 27 ≤ 30 ✓ (true). Checking x + y ≥ 18: 9 + 9 = 18 ≥ 18 ✓ (true). Checking x ≤ 10: 9 ≤ 10 ✓ (true). Checking x ≥ 0: 9 ≥ 0 ✓ and y ≥ 0: 9 ≥ 0 ✓ (true). Also checking context: both 9 and 9 are nonnegative whole numbers ✓. Conclusion: viable because all constraints are satisfied. Choice A is correct because it properly checks all constraints showing (9, 9) is viable—all inequalities are satisfied and the values make sense in context. Choice B incorrectly calculates: it says 2(9) + 9 = 27, which is correct, but then claims this violates the constraint when actually 27 ≤ 30 is true! When checking inequalities, be careful with the direction: ≤ means 'less than or equal to,' so 27 ≤ 30 is satisfied. The viability-checking procedure: Make a checklist of every constraint. For each one, substitute the point and check if it's satisfied. Write 'Yes' or 'No' next to each constraint. If even one 'No' appears, the solution is nonviable—identify which constraint(s) failed. Also do a reality check: negative quantities? fractional items when must be whole? These context violations also make solutions nonviable!

Question 9

1. $\{2x+y\le 50,\; x+y\le 30,\; x\ge 8,\; x\ge 0,\; y\ge 0\}$ (correct answer)
2. $\{2x+y\le 50,\; x+y\ge 30,\; x\ge 8,\; x\ge 0,\; y\ge 0\}$
3. $\{2x+y\ge 50,\; x+y\le 30,\; x\ge 8,\; x\ge 0,\; y\ge 0\}$
4. $\{2x+y\le 50,\; x+y\le 30,\; x\le 8,\; x\ge 0,\; y\ge 0\}$

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. When setting up constraints from context, look for key phrases: 'at most' or 'no more than' → ≤, 'at least' or 'minimum' → ≥, 'exactly' or 'must be' → =, 'less than' → <, 'more than' → >. Also, don't forget implicit constraints like x ≥ 0 and y ≥ 0 (can't have negative quantities) or x, y must be integers (if counting discrete items). Let's identify all the constraints from the farming situation: water 'at most 50 units' gives us 2x + y ≤ 50, total acres 'at most 30' gives us x + y ≤ 30, minimum corn 'at least 8 acres' gives us x ≥ 8, and non-negativity x ≥ 0, y ≥ 0. Putting this together as a system: {2x + y ≤ 50, x + y ≤ 30, x ≥ 8, x ≥ 0, y ≥ 0}. This system captures all the limitations and requirements of the situation. Choice A is correct because it includes all constraints with correct inequality directions. Choice C has the inequality direction backwards: 'at most 30 acres' means ≤, not ≥. This is a super common error! When total acres 'at most 30,' we need x + y ≤ 30, meaning we can't exceed 30. The 'at most' puts a ceiling, not a floor. The constraint-writing recipe: (1) List EVERY limitation mentioned in the problem (budget, time, capacity, minimums, etc.), (2) For each, identify the inequality symbol from key words ('at most' → ≤, 'at least' → ≥, etc.), (3) Write the mathematical inequality using the costs, rates, or quantities from context, (4) Don't forget implicit constraints like x ≥ 0, y ≥ 0 (non-negativity) or x, y integers (if discrete). Make sure your system captures EVERY constraint!

Question 10

1. It violates only $3x+2y\le 24$. (correct answer)
2. It violates only $x+y\le 10$.
3. It violates both $3x+2y\le 24$ and $x+y\le 10$.
4. It violates no constraints; the point is viable.

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. Checking viability is systematic: (1) Substitute the proposed solution into each constraint, (2) Verify each inequality or equation is satisfied, (3) Check context realism (non-negative? whole numbers if needed?). If everything checks out, it's viable. If anything fails, it's nonviable—and you should explain which constraint was violated. To check if (6, 4) is viable, we substitute into each constraint: Flour constraint: 3(6) + 2(4) = 18 + 8 = 26 ≤ 24? NO! 26 > 24, so this constraint is violated. Oven capacity: 6 + 4 = 10 ≤ 10? YES! This constraint is satisfied. Minimum muffins: 6 ≥ 2? YES! This constraint is satisfied. Non-negativity: 6 ≥ 0 and 4 ≥ 0? YES! Both satisfied. Conclusion: nonviable because the flour constraint is violated. Choice A is correct because it correctly identifies that only the flour constraint 3x + 2y ≤ 24 is violated by the point (6, 4). Choice C says both constraints are violated, but we verified that x + y = 10 ≤ 10, so the oven capacity constraint is actually satisfied. One failure disqualifies the entire solution, but it's important to accurately identify which specific constraints are violated! When checking viability, substitute carefully: if the point is (6, 4), that means x = 6 and y = 4. Substitute those values into EVERY inequality and equation. For example, if one constraint is 3x + 2y ≤ 24, check: 3(6) + 2(4) = 18 + 8 = 26 ≤ 24? No! Do this for every single constraint. It's tedious but necessary—one missed check could mean accepting an infeasible solution!

Question 11

1. $\{3x+y\le 18,\ x+y\ge 10,\ x\ge 0,\ y\ge 0\}$
2. $\{3x+y=18,\ x+y\ge 10,\ x\ge 0,\ y\ge 0\}$ (correct answer)
3. $\{3x+y=18,\ x+y\le 10,\ x\ge 0,\ y\ge 0\}$
4. $\{x+3y=18,\ x+y\ge 10,\ x\ge 0,\ y\ge 0\}$

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. When setting up constraints from context, look for key phrases: 'at most' or 'no more than' → ≤, 'at least' or 'minimum' → ≥, 'exactly' or 'must be' → =, 'less than' → <, 'more than' → >. Also, don't forget implicit constraints like x ≥ 0 and y ≥ 0 (can't have negative quantities) or x, y must be integers (if counting discrete items). Let's identify all the constraints from 'buying notebooks and pens': The cost constraint 'has exactly $18 to spend' gives us 3x + y = 18 (note the equals sign for 'exactly'!). The quantity constraint 'wants at least 10 items total' gives us x + y ≥ 10. The non-negativity constraints give us x ≥ 0 and y ≥ 0. Putting this together as a system: {3x + y = 18, x + y ≥ 10, x ≥ 0, y ≥ 0}. This system captures all the limitations and requirements of the situation. Choice B is correct because it includes all constraints with correct inequality directions and uses = for the 'exactly$18' constraint. Choice A uses an inequality (≤) when the context describes an equation (=): 'exactly $18' indicates the student must spend all$18, not just up to $18. Reserve = for 'exactly,' 'must be,' or 'equal to' situations. The word 'exactly' is your clue that this isn't a maximum or minimum—it's a precise requirement! The constraint-writing recipe: (1) List EVERY limitation mentioned in the problem (budget, time, capacity, minimums, etc.), (2) For each, identify the inequality symbol from key words ('at most' → ≤, 'at least' → ≥, etc.), (3) Write the mathematical inequality using the costs, rates, or quantities from context, (4) Don't forget implicit constraints like x ≥ 0, y ≥ 0 (non-negativity) or x, y integers (if discrete). Make sure your system captures EVERY constraint!

Question 12

1. Viable: it satisfies both $15x+40y\le 400$ and $x+y\le 20$.
2. Nonviable: it violates the budget constraint since $15(12)+40(6)=420>400$. (correct answer)
3. Nonviable: it violates the storage constraint since $12+6=20>20$.
4. Nonviable: it violates non-negativity because $y$ must be negative to meet the budget.

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. Checking viability is systematic: (1) Substitute the proposed solution into each constraint, (2) Verify each inequality or equation is satisfied, (3) Check context realism (non-negative? whole numbers if needed?). If everything checks out, it's viable. If anything fails, it's nonviable—and you should explain which constraint was violated. To check if (12, 6) is viable, we substitute into each constraint: for budget 15x + 40y ≤ 400, 15(12) + 40(6) = 180 + 240 = 420 ≤ 400? (false, 420 > 400); for storage x + y ≤ 20, 12 + 6 = 18 ≤ 20 (true). Also checking context: x=12≥0, y=6≥0, whole numbers (yes). Conclusion: nonviable because budget constraint violated. Choice B is correct because it correctly identifies the violation in the budget constraint. Choice A says the point is viable but doesn't check all constraints: it claims satisfaction but misses that 420 > 400 violates budget. To determine viability, you MUST check every single constraint—missing even one can lead to accepting an impossible solution! The viability-checking procedure: Make a checklist of every constraint. For each one, substitute the point and check if it's satisfied. Write 'Yes' or 'No' next to each constraint. If even one 'No' appears, the solution is nonviable—identify which constraint(s) failed. Also do a reality check: negative quantities? fractional items when must be whole? These context violations also make solutions nonviable!

Question 13

1. Nonviable because it violates $2x+y\le 40$.
2. Viable because it satisfies all three constraints.
3. Nonviable because it violates $x\ge 8$.
4. Nonviable because it violates $x+y\le 30$. (correct answer)

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. Checking viability is systematic: (1) Substitute the proposed solution into each constraint, (2) Verify each inequality or equation is satisfied, (3) Check context realism (non-negative? whole numbers if needed?). If everything checks out, it's viable. If anything fails, it's nonviable—and you should explain which constraint was violated. To check if (12, 20) is viable, we substitute into each constraint: Total plants: x + y = 12 + 20 = 32 > 30 ✗. Stakes needed: 2(12) + 20 = 24 + 20 = 44 > 40 ✗. Minimum tomatoes: x = 12 ≥ 8 ✓. Non-negativity: x = 12 ≥ 0 ✓ and y = 20 ≥ 0 ✓. Conclusion: nonviable because it violates both the total plants constraint (32 > 30) and the stakes constraint (44 > 40). Choice B is correct because it correctly identifies that the solution violates x + y ≤ 30. Choice C says it violates only the stakes constraint, but actually (12, 20) violates BOTH the total plants constraint (32 > 30) AND the stakes constraint (44 > 40). When a solution violates multiple constraints, it's important to identify the primary or most obvious violation—here, exceeding the garden's 30-plant capacity is the clearest issue! When checking viability, substitute carefully: if the point is (12, 20), that means x = 12 and y = 20. Substitute those values into EVERY inequality and equation. For example, if one constraint is x + y ≤ 30, check: 12 + 20 = 32 ≤ 30? No! Do this for every single constraint. It's tedious but necessary—one missed check could mean accepting an infeasible solution!

Question 14

1. Viable because protein is $8(3)+12(2)=48\ge 48$ and calories are $200(3)+150(2)=900\le 900$. (correct answer)
2. Nonviable because protein is $8(3)+12(2)=44<48$.
3. Nonviable because calories are $200(3)+150(2)=950>900$.
4. Viable because calories are under 900 even though protein is below 48.

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. Checking viability is systematic: (1) Substitute the proposed solution into each constraint, (2) Verify each inequality or equation is satisfied, (3) Check context realism (non-negative? whole numbers if needed?). If everything checks out, it's viable. If anything fails, it's nonviable—and you should explain which constraint was violated. To check if (3, 2) is viable, we substitute into each constraint: Protein constraint: 8(3) + 12(2) = 24 + 24 = 48 ≥ 48 ✓. Calorie constraint: 200(3) + 150(2) = 600 + 300 = 900 ≤ 900 ✓. Non-negativity: x = 3 ≥ 0 ✓ and y = 2 ≥ 0 ✓. Conclusion: viable because all constraints are satisfied. Choice A is correct because it properly checks all constraints showing (3, 2) is viable. Choice B says the point is nonviable but makes a calculation error: 8(3) + 12(2) = 24 + 24 = 48, not 44. The protein requirement of exactly 48g is met! This highlights how arithmetic errors can lead to incorrect viability conclusions—always double-check your calculations! When checking viability, substitute carefully: if the point is (3, 2), that means x = 3 and y = 2. Substitute those values into EVERY inequality and equation. For example, if one constraint is 8x + 12y ≥ 48, check: 8(3) + 12(2) = 24 + 24 = 48 ≥ 48? Yes! Do this for every single constraint. It's tedious but necessary—one missed check could mean accepting an infeasible solution!

Question 15

1. (i) viable, (ii) viable, (iii) nonviable
2. (i) viable, (ii) nonviable, (iii) viable
3. (i) nonviable, (ii) viable, (iii) viable (correct answer)
4. (i) viable, (ii) nonviable, (iii) nonviable

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. A solution is viable (feasible) if it satisfies EVERY SINGLE constraint AND makes sense in the real-world context (like no negative quantities, whole items when you can't buy half an item, etc.). If even one constraint is violated, or if the solution is unrealistic, it's nonviable. Think of constraints like security checkpoints—you need to pass through all of them! Let's check each point against all constraints: For (6, 9): 3(6) + 5(9) = 18 + 45 = 63 > 60 ✗ → nonviable because it exceeds board limit. For (8, 6): 3(8) + 5(6) = 24 + 30 = 54 ≤ 60 ✓, 8 + 6 = 14 ≤ 18 ✓, x = 8 ≥ 6 ✓, x = 8 ≥ 0 ✓, y = 6 ≥ 0 ✓ → viable. For (12, 2): 3(12) + 5(2) = 36 + 10 = 46 ≤ 60 ✓, 12 + 2 = 14 ≤ 18 ✓, x = 12 ≥ 6 ✓, x = 12 ≥ 0 ✓, y = 2 ≥ 0 ✓ → viable. Notice how (6, 9) fails the board constraint, while (8, 6) and (12, 2) pass all constraints. Even one violation makes a solution nonviable! Choice A is correct because it correctly identifies (i) as nonviable and (ii) and (iii) as viable. Choice B incorrectly says (iii) is nonviable, but for (12, 2), all constraints are satisfied: 3(12) + 5(2) = 46 ≤ 60 ✓, 12 + 2 = 14 ≤ 18 ✓, and 12 ≥ 6 ✓. This point is perfectly viable! Always check your arithmetic carefully when evaluating multiple points. The viability-checking procedure: Make a checklist of every constraint. For each one, substitute the point and check if it's satisfied. Write 'Yes' or 'No' next to each constraint. If even one 'No' appears, the solution is nonviable—identify which constraint(s) failed. Also do a reality check: negative quantities? fractional items when must be whole? These context violations also make solutions nonviable!

Question 16

1. It violates $2x+3y\le 24$.
2. It violates $x+y\ge 4$.
3. It violates both $2x+3y\le 24$ and $x+y\ge 4$.
4. It violates no constraints (it is viable). (correct answer)

Explanation: This question tests your ability to translate real-world constraints into mathematical inequalities (and equations) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in the context. Checking viability is systematic: (1) Substitute the proposed solution into each constraint, (2) Verify each inequality or equation is satisfied, (3) Check context realism (non-negative? whole numbers if needed?). If everything checks out, it's viable. If anything fails, it's nonviable—and you should explain which constraint was violated. To check if (3, 2) is viable, we substitute into each constraint: Instructor time constraint: 2(3) + 3(2) = 6 + 6 = 12 ≤ 24 ✓ True. Minimum classes constraint: 3 + 2 = 5 ≥ 4 ✓ True. Non-negativity: x = 3 ≥ 0 ✓ and y = 2 ≥ 0 ✓. Conclusion: viable because all constraints are satisfied. Choice D is correct because it correctly identifies that the point (3, 2) satisfies all constraints and is therefore viable. Choice A is missing a crucial constraint check: it claims the point violates 2x + 3y ≤ 24, but 2(3) + 3(2) = 12 ≤ 24 is true! This shows the importance of careful calculation when checking constraints. When checking viability, substitute carefully: if the point is (3, 2), that means x = 3 and y = 2. Substitute those values into EVERY inequality and equation. For example, if one constraint is 2x + 3y ≤ 24, check: 2(3) + 3(2) = 6 + 6 = 12 ≤ 24? Yes! Do this for every single constraint. It's tedious but necessary—one missed check could mean accepting an infeasible solution!

Question 17

1. $15x \geq 300 + 8x$ and $x \geq 0$ (correct answer)
2. $15x > 300 + 8x$ and $x > 0$
3. $15x \geq 300 + 8$ and $x \geq 0$
4. $15x - 8x \geq 300$ and $x \geq 0$

Explanation: To break even or make a profit, revenue ($15x$) must be greater than or equal to total costs ($300 + 8x$). The number of tickets sold cannot be negative. Choice A correctly represents both constraints. Choice B uses strict inequality when break-even should be included. Choice C omits the variable cost per person. Choice D is algebraically equivalent to A but doesn't clearly show the relationship between revenue and costs.

Question 18

1. $-2x^2 + 16x - 24 \geq 8000$; $x = 2$ is not viable since units are inconsistent
2. $-2x^2 + 16x - 24 \geq 8$; $x = 2$ is not viable since it yields $0 profit (correct answer)
3. $-2x^2 + 16x - 24 \geq 8$; $x = 2$ is viable since it equals minimum requirement
4. $-2x^2 + 16x - 24 > 8$; $x = 2$ is not viable since strict inequality excludes boundary

Explanation: When you encounter profit inequality problems, pay close attention to units and carefully match the constraint language to the mathematical expression. The profit function gives you $P$ in thousands of dollars, so you need to convert the $8,000 requirement to the same units. Since $P$ represents profit in thousands of dollars, a profit of "at least 8,000" means $$P \geq 8$$ (because 8,000 = 8 thousands). This gives you the inequality $-2x^2 + 16x - 24 \geq 8$. Now test $x = 2$: $P = -2(2)^2 + 16(2) - 24 = -8 + 32 - 24 = 0$. Since 0 is not greater than or equal to 8, $x = 2$ doesn't meet the profit requirement. Looking at the wrong answers: Choice A incorrectly uses 8000 instead of 8, ignoring that $P$ is already in thousands. This creates a units mismatch that makes the inequality impossible to satisfy. Choice C correctly sets up the inequality but wrongly claims $x = 2$ is viable—substituting shows it yields 0, which doesn't meet the "at least 8" requirement. Choice D uses a strict inequality ($>$) instead of "greater than or equal to" ($\geq$), which doesn't match the phrase "at least." The correct answer is B because it properly converts units and correctly evaluates that $x = 2$ yields insufficient profit. Study tip: Always check that your units are consistent throughout the problem, and remember that "at least" translates to $\geq$, not $>$.

Question 19

1. $3s + 6\ell \leq 48$ (trim constraint)
2. $2s + 4\ell \leq 20$ (plywood constraint)
3. $s \geq 0$ and $\ell \geq 0$ (non-negativity)
4. $s + \ell \geq 1$ (production requirement) (correct answer)

Explanation: Choices A, B, and C correctly represent the material constraints and logical requirements (can't make negative shelves). Choice D assumes a production requirement that wasn't stated in the problem. The carpenter could choose to make zero shelves if desired, making this constraint invalid. This tests the ability to distinguish between given constraints and assumed constraints.

Question 20

1. $r + p \leq 100$, $8r + 12p - 3r - 5p \geq 200$, $r \geq 0$, $p \geq 0$
2. $r + p \leq 100$, $8r + 12p \geq 200$, $r \geq 0$, $p \geq 0$
3. $r + p \leq 100$, $(8-3)r + (12-5)p \geq 200$, $r \geq 0$, $p \geq 0$ (correct answer)
4. $r + p = 100$, $5r + 7p \geq 200$, $r > 0$, $p > 0$

Explanation: The constraints are: (1) At most 100 meals: $r + p \leq 100$, (2) At least $200 profit: profit per regular meal is $8 - 3 = 5$, profit per premium meal is $12 - 5 = 7$, so $5r + 7p \geq 200$, (3) Non-negative production: $r \geq 0, p \geq 0$. Choice C shows the profit calculation explicitly. Choice A shows the calculation but doesn't simplify. Choice B uses revenue instead of profit. Choice D requires exactly 100 meals and positive production, which aren't required.