Modeling with Equation/InequalityConstraints - Algebra 1

A community garden is planting vegetables. Let $x$ be the number of tomato plants and $y$ be the number of pepper plants. Constraints: - There is space for at most 40 plants total: $x+y\le 40$ . - Tomato plants must be at least 10: $x\ge 10$ . - Pepper plants cannot exceed 25: $y\le 25$ . - $x$ and $y$ must be nonnegative whole numbers. Is the point $(x,y)=(12,28)$ a viable solution? Explain by identifying any constraint it violates.
Answer: Nonviable; it violates $y\le 25$ because $28>25$ . 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, 28) is viable, we substitute into each constraint: Checking x + y ≤ 40: 12 + 28 = 40 ≤ 40 ✓ (true). Checking x ≥ 10: 12 ≥ 10 ✓ (true). Checking y ≤ 25: 28 ≤ 25 ✗ (false, since 28 > 25). Also checking context: both 12 and 28 are nonnegative whole numbers ✓. Conclusion: nonviable because it violates the constraint y ≤ 25. Choice C is correct because it correctly identifies the violation in the constraint y ≤ 25, since 28 > 25. Choice A says the point is viable but doesn't check all constraints: it misses that 28 > 25 violates the pepper plant limit. 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!
If $r$ represents regular meals and $p$ represents premium meals, which system correctly models all constraints for viable daily operation?
Answer: $r + p \leq 100$ , $(8-3)r + (12-5)p \geq 200$ , $r \geq 0$ , $p \geq 0$ 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.
A school club is ordering snacks for an event. Let $x$ be the number of boxes of granola bars and $y$ be the number of boxes of fruit snacks. Constraints: - Each granola box costs $6 and each fruit snack box costs $4. - The club can spend at most $60. - They need at least 12 boxes total. - They cannot order a negative number of boxes. Which system of inequalities models this situation?
Answer: $\{6x+4y\le 60,\ x+y\ge 12,\ x\ge 0,\ y\ge 0\}$ 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 'ordering snacks for an event': The cost constraint 'can spend at most $60' gives us 6x + 4y ≤ 60. The quantity constraint 'need at least 12 boxes total' gives us x + y ≥ 12. The non-negativity constraints 'cannot order a negative number of boxes' give us x ≥ 0 and 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 A has the inequality directions 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!
A small theater sells adult and student tickets. Let $x$ be the number of adult tickets and $y$ be the number of student tickets. Constraints: - The theater has 120 seats: $x+y\le 120$ . - Adult tickets cost $10 and student tickets cost $6, and the theater wants at least $900 in revenue: $10x+6y\ge 900$ . - $x\ge 0$ and $y\ge 0$ . Which points represent feasible (viable) solutions?
Answer: Both $(60,60)$ and $(50,70)$ 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 (60, 60): x + y = 60 + 60 = 120 ≤ 120 ✓, 10(60) + 6(60) = 600 + 360 = 960 ≥ 900 ✓, x = 60 ≥ 0 ✓, y = 60 ≥ 0 ✓ → viable. For (50, 70): x + y = 50 + 70 = 120 ≤ 120 ✓, 10(50) + 6(70) = 500 + 420 = 920 ≥ 900 ✓, x = 50 ≥ 0 ✓, y = 70 ≥ 0 ✓ → viable. For (30, 90): x + y = 30 + 90 = 120 ≤ 120 ✓, 10(30) + 6(90) = 300 + 540 = 840 < 900 ✗ → nonviable because revenue is below minimum. Notice how (60, 60) and (50, 70) pass all constraints, while (30, 90) fails the revenue constraint. Even one violation makes a solution nonviable! Choice D is correct because both (60, 60) and (50, 70) satisfy all constraints. Choice C incorrectly identifies only (30, 90) as viable, but this point violates the revenue constraint: 10(30) + 6(90) = 840 < 900. The theater needs at least $900 in revenue, and $840 falls short! This demonstrates why checking ALL constraints systematically is crucial—a point might satisfy some constraints while violating others. 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!
A bakery makes muffins and cookies. Let $x$ be the number of muffin batches and $y$ be the number of cookie batches. Constraints: - Each muffin batch takes 2 hours and each cookie batch takes 3 hours. - The bakery has at most 18 hours available. - Each muffin batch uses 4 cups of flour and each cookie batch uses 2 cups of flour. - The bakery has at most 24 cups of flour. - $x\ge 0$ and $y\ge 0$ . Which constraints does the point $(x,y)=(6,2)$ violate?
Answer: It violates only the flour constraint. 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, 2) is viable, we substitute into each constraint: Time constraint: 2(6) + 3(2) = 12 + 6 = 18 ≤ 18 ✓. Flour constraint: 4(6) + 2(2) = 24 + 4 = 28 > 24 ✗. Non-negativity: x = 6 ≥ 0 ✓ and y = 2 ≥ 0 ✓. Conclusion: nonviable because the flour constraint is violated (28 > 24). Choice B is correct because it correctly identifies the violation in the flour constraint. Choice D says the point is viable but doesn't check all constraints: it misses that 4(6) + 2(2) = 28 > 24. 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 (6, 2), that means x = 6 and y = 2. Substitute those values into EVERY inequality and equation. For example, if one constraint is 4x + 2y ≤ 24, check: 4(6) + 2(2) = 24 + 4 = 28 ≤ 24? No! Do this for every single constraint. It's tedious but necessary—one missed check could mean accepting an infeasible solution!
A theater is deciding how many adult and student tickets to sell for a fundraiser. Let $x$ be the number of adult tickets and $y$ be the number of student tickets. - The theater can sell at most 300 tickets: $x+y \le 300$ . - Adult tickets are $\12$ and student tickets are $\8$ . The fundraiser must bring in at least $\2400$ : $12x+8y \ge 2400$ . - At least 50 student tickets must be sold: $y \ge 50$ . - $x \ge 0$ and $y \ge 0$ . What system of constraints models this situation?
Answer: $\{x+y \le 300,\ 12x+8y \ge 2400,\ y \ge 50,\ x \ge 0,\ y \ge 0\}$ 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 $x \ge 0$ and $y \ge 0$ (can't have negative quantities) or x, y must be integers (if counting discrete items). The constraints are: (1) total tickets 'at most 300' translates to $x + y \le 300$ because it's a maximum capacity, (2) revenue 'at least $2400' translates to $12x + 8y \ge 2400$ because it's a minimum requirement, (3) 'at least 50 student tickets' translates to $y \ge 50$ because of the minimum, and implicit non-negativity $x \ge 0$ , $y \ge 0$ because you can't sell negative tickets. Together: { $x + y \le 300$ , $12x + 8y \ge 2400$ , $y \ge 50$ , $x \ge 0$ , $y \ge 0$ }. Any solution in this system's feasible region is a viable option for the context. Choice B is correct because it includes all constraints with correct inequality directions. Choice C has the inequality direction backwards: 'at least $2400' means ≥ 2400, not ≤, which would incorrectly cap revenue instead of requiring a minimum. This is a super common error! When revenue 'at least $2400,' we need $12x + 8y \ge 2400$ , meaning we can't fall below. 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 $x \ge 0$ , $y \ge 0$ (non-negativity) or x, y integers (if discrete). Make sure your system captures EVERY constraint!
A school club is buying snacks for an event. Let $x$ be the number of boxes of granola bars and $y$ be the number of boxes of fruit snacks. Each granola bar box costs $6 and each fruit snack box costs $4. The club has at most $60 to spend and needs at least 12 total boxes. Also, $x$ and $y$ cannot be negative. Which system of inequalities models these constraints?
Answer: $\{6x+4y\le 60,\ x+y\ge 12,\ x\ge 0,\ y\ge 0\}$ 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!
A gardener is planting tomatoes and peppers. Let $x$ be the number of tomato plants and $y$ be the number of pepper plants. Constraints: - The garden can hold at most 30 plants total: $x+y\le 30$ . - Tomato plants require 2 stakes each and pepper plants require 1 stake each. The gardener has at most 40 stakes: $2x+y\le 40$ . - The gardener wants at least 8 tomato plants: $x\ge 8$ . - $x\ge 0$ and $y\ge 0$ . Is the point $(x,y)=(12,20)$ a viable solution?
Answer: Nonviable because it violates $x+y\le 30$ . 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!
A carpenter has 48 feet of trim and 20 square feet of plywood. Small shelves use 3 feet of trim and 2 square feet of plywood. Large shelves use 6 feet of trim and 4 square feet of plywood. If the carpenter makes $s$ small shelves and $\ell$ large shelves, which constraint is NOT properly represented?
Answer: $s + \ell \geq 1$ (production requirement) 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.
A community garden is planning plots for tomatoes and peppers. Let $x$ be the number of tomato plots and $y$ be the number of pepper plots. - There are at most 30 plots total: $x+y\le 30$ . - Tomato plots must be at least 8: $x\ge 8$ . - Each tomato plot needs 3 bags of soil and each pepper plot needs 2 bags. The garden has at most 70 bags of soil: $3x+2y\le 70$ . - $x\ge 0$ and $y\ge 0$ . Is the point $(x,y)=(10,12)$ a viable solution? Explain by checking the constraints.
Answer: Viable; it satisfies $x+y\le 30$ , $x\ge 8$ , $3x+2y\le 70$ , and $x,y\ge 0$ . 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 (10, 12) is viable, we substitute into each constraint: for x + y ≤ 30, 10 + 12 = 22 ≤ 30 (true); for x ≥ 8, 10 ≥ 8 (true); for 3x + 2y ≤ 70, 3(10) + 2(12) = 30 + 24 = 54 ≤ 70 (true); for x ≥ 0 and y ≥ 0 (true). Also checking context: non-negative and whole numbers, yes. Conclusion: viable because all satisfied. Choice A is correct because it properly checks all constraints showing (10, 12) is viable. Choice B says the point is nonviable but doesn't check all constraints: it only checks the soil but with a calculation error (54, not 74), missing that it actually satisfies 54 ≤ 70. 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!
A pet shelter is buying bags of dog food and cat food. Let $x$ be the number of dog-food bags and $y$ be the number of cat-food bags. - Dog food costs $18 per bag and cat food costs $12 per bag. Budget is at most $240: $18x+12y\le 240$ . - Storage space allows at most 20 bags total: $x+y\le 20$ . - They need at least 6 dog-food bags: $x\ge 6$ . - $x\ge 0$ and $y\ge 0$ . Check whether $(x,y)=(6,12)$ satisfies all constraints.
Answer: Nonviable; it violates the budget constraint because $18(6)+12(12)=252>240$ . 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 (6, 12) is viable, we substitute into each constraint: for 18x + 12y ≤ 240, 18(6) + 12(12) = 108 + 144 = 252 > 240 (false); for x + y ≤ 20, 6 + 12 = 18 ≤ 20 (true); for x ≥ 6, 6 ≥ 6 (true); for x ≥ 0, y ≥ 0 (true). Also checking context: non-negative and whole numbers, yes. Conclusion: nonviable because budget violated. Choice B is correct because it correctly identifies the violation in the budget constraint. Choice C says it's nonviable but incorrectly identifies the violation: it checks storage as 22 > 20, but actually 18 ≤ 20 is satisfied. If ANY constraint is violated, the solution is nonviable—period. One failure disqualifies the entire solution, no matter how many other constraints it satisfies! 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!
A workshop produces birdhouses and planters. Let $x$ be the number of birdhouses and $y$ be the number of planters. Constraints: - Each birdhouse uses 3 boards and each planter uses 5 boards. There are at most 60 boards: $3x+5y\le 60$ . - The workshop can make at most 18 total items: $x+y\le 18$ . - Production requires at least 6 birdhouses: $x\ge 6$ . - $x\ge 0$ and $y\ge 0$ . Determine if each solution is viable or nonviable: (i) $(6,9)$ , (ii) $(8,6)$ , (iii) $(12,2)$ .
Answer: (i) nonviable, (ii) viable, (iii) viable 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!
A company's profit $P$ (in thousands) is modeled by $P = -2x^2 + 16x - 24$ , where $x$ is the number of products sold (in hundreds). The company needs a profit of at least $8,000. Which represents this constraint and identifies whether $x = 2$ is a viable option?
Answer: $-2x^2 + 16x - 24 \geq 8$ ; $x = 2$ is not viable since it yields $0 profit 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 $>$ .
A farmer plants acres of corn and acres of soybeans. Corn requires 2 units of water per acre and soybeans require 1 unit of water per acre. The farmer has at most 50 units of water and can plant at most 30 acres total. Also, the farmer must plant at least 8 acres of corn. Let $x$ = acres of corn and $y$ = acres of soybeans. What system of constraints models the situation?
Answer: $\{2x+y\le 50,\; x+y\le 30,\; x\ge 8,\; x\ge 0,\; y\ge 0\}$ 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!
A theater is choosing how many balcony seats and floor seats to sell. Let $x$ = balcony tickets and $y$ = floor tickets. The theater can sell at most 300 total tickets: $x+y\le 300$ . Fire code requires at least 120 floor tickets to be sold for staffing plans: $y\ge 120$ . Also, $x\ge 0$ and $y\ge 0$ . Which point is nonviable?
Answer: $(210,110)$ 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!
A student is buying notebooks and pens. Let $x$ be the number of notebooks and $y$ be the number of pens. - Notebooks cost $4 each and pens cost $1 each. The student has at most $25: $4x+y\le 25$ . - The student needs at least 8 items total: $x+y\ge 8$ . - Notebooks and pens must be whole numbers (you can't buy a fraction of an item), and $x,y\ge 0$ . Which statement is true about the point $(x,y)=(3.5,6)$ ?
Answer: Nonviable; it violates the whole-number requirement even though it satisfies the inequalities. 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!
A smoothie shop is mixing two bases. Let $x$ be the number of scoops of yogurt base and $y$ be the number of scoops of juice base. Constraints: - The smoothie must contain exactly 10 scoops total: $x+y=10$ . - For taste, at least 4 scoops must be yogurt: $x\ge 4$ . - For sugar limits, at most 7 scoops can be juice: $y\le 7$ . - $x\ge 0$ and $y\ge 0$ . Is $(x,y)=(3,7)$ a viable solution?
Answer: Nonviable; it violates $x\ge 4$ because $3<4$ . 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, 7) is viable, we substitute into each constraint: Checking x + y = 10: 3 + 7 = 10 ✓ (true, the equation is satisfied). Checking x ≥ 4: 3 ≥ 4? No, 3 < 4 ✗ (false). Checking y ≤ 7: 7 ≤ 7 ✓ (true, equal satisfies ≤). Checking x ≥ 0: 3 ≥ 0 ✓ and y ≥ 0: 7 ≥ 0 ✓ (true). Conclusion: nonviable because it violates the constraint x ≥ 4 (need at least 4 scoops of yogurt, but only have 3). Choice B is correct because it correctly identifies the violation in the constraint x ≥ 4, since 3 < 4. Choice A says the point is viable but doesn't check all constraints: it misses that 3 < 4 violates the minimum yogurt requirement. To determine viability, you MUST check every single constraint—missing even one can lead to accepting an impossible solution! 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!
A car rental company has two vehicle types: compact and SUV. Let $x$ be the number of compact cars rented and $y$ be the number of SUVs rented. Constraints: - There are at most 25 vehicles available: $x+y\le 25$ . - Compact rentals bring in $30 each and SUV rentals bring in $50 each. The company needs at least $900 in revenue: $30x+50y\ge 900$ . - $x\ge 0$ and $y\ge 0$ . Which constraints does the point $(x,y)=(10,10)$ violate?
Answer: It violates only the revenue constraint $30x+50y\ge 900$ . 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 (10, 10) is viable, we substitute into each constraint: Vehicle availability: x + y = 10 + 10 = 20 ≤ 25 ✓. Revenue requirement: 30(10) + 50(10) = 300 + 500 = 800 < 900 ✗. Non-negativity: x = 10 ≥ 0 ✓ and y = 10 ≥ 0 ✓. Conclusion: nonviable because the revenue constraint is violated (800 < 900). Choice B is correct because it correctly identifies that the solution violates only the revenue constraint 30x + 50y ≥ 900. Choice D says the point is viable but doesn't check all constraints: it misses that 30(10) + 50(10) = 800 < 900. The company needs at least $900 in revenue, but this combination only generates $800. 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 (10, 10), that means x = 10 and y = 10. Substitute those values into EVERY inequality and equation. For example, if one constraint is 30x + 50y ≥ 900, check: 30(10) + 50(10) = 300 + 500 = 800 ≥ 900? No! Do this for every single constraint. It's tedious but necessary—one missed check could mean accepting an infeasible solution!
A theater is choosing how many adult and student tickets to sell in advance for a small show. Let $x$ = number of adult tickets and $y$ = number of student tickets. Constraints: - The theater has at most 120 seats: $x+y\le 120$ . - Adult tickets are $12 and student tickets are $8; advance sales revenue must be at least $1000: $12x+8y\ge 1000$ . - At least 30 student tickets must be available: $y\ge 30$ . - $x\ge 0$ and $y\ge 0$ . Is the point $(70, 40)$ a viable solution?
Answer: Viable; it meets the seat limit, revenue requirement, and minimum student tickets. 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 (70, 40) is viable, we substitute into each constraint: Seat limit: 70 + 40 = 110 ≤ 120? YES! They're using 110 of 120 available seats. Revenue requirement: 12(70) + 8(40) = 840 + 320 = 1160 ≥ 1000? YES! Revenue of $1160 exceeds the $1000 minimum. Minimum student tickets: 40 ≥ 30? YES! They have 40 student tickets, exceeding the minimum of 30. Non-negativity: 70 ≥ 0 and 40 ≥ 0? YES! Both are non-negative. Conclusion: viable because all constraints are satisfied. Choice A is correct because it properly checks all constraints showing (70, 40) is viable: seat limit satisfied, revenue requirement met, and minimum student tickets achieved. Choice B incorrectly claims the seat limit is violated because 70 + 40 > 120, but 70 + 40 = 110, which is less than 120. Accurate arithmetic is crucial when checking constraints! 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. Remember to check your arithmetic carefully—a calculation error could lead to incorrectly accepting or rejecting a solution!
A community center is scheduling yoga classes and art classes. Let $x$ = number of yoga classes and $y$ = number of art classes. Each yoga class uses 2 hours of instructor time and each art class uses 3 hours. There are at most 24 hours of instructor time available. The center must offer at least 4 total classes. Also, $x\ge 0$ and $y\ge 0$ . Which constraints does the point $(3,2)$ violate, if any?
Answer: It violates no constraints (it is viable). 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!

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