Modeling with Equation/InequalityConstraints - Algebra 2

A movie theater sells adult tickets and student tickets. Let $a$ = number of adult tickets and $s$ = number of student tickets. Adult tickets cost $12 and student tickets cost $8. The theater wants revenue of at least $480 from a show, but can sell at most 50 tickets total. Tickets must be whole numbers and nonnegative. Which constraint system represents this situation?
Answer: $\begin{cases}12a+8s\ge 480\\a+s\le 50\\a\ge 0,\\ s\ge 0\\a,s\in\mathbb{Z}\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! The constraints are revenue $12a + 8s \ge 480$ (at least $480$ ), total $a + s \le 50$ (at most 50), and $a \ge 0$ , $s \ge 0$ with integers (nonnegative whole tickets). Choice B correctly represents all constraints with ≥ for revenue and ≤ for total, including nonnegativity and integers. A distractor like Choice C changes total to ≥50, but it's at most 50—pay attention to 'at most' versus 'at least' in the context! Constraint identification from context: (1) list every limitation mentioned ('budget $X$ ,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like $x \ge 0$ , $y \ge 0$ (can't be negative) or $x$ , $y$ integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A school is buying notebooks and folders. Let $x$ = number of notebooks and $y$ = number of folders. Constraints: - Notebooks cost $2 each and folders cost $3 each; the budget is at most $60: $2x+3y\le 60$ . - They need at least 10 notebooks: $x\ge 10$ . - They need at least 8 folders: $y\ge 8$ . - Items must be whole numbers and non-negative. Which system correctly represents all constraints?
Answer: $\begin{cases}2x+3y\le 60\\x\ge 10\\y\ge 8\\x\ge 0,\\ y\ge 0\\x,y\in\mathbb{Z}\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! Let's translate each constraint: (1) Budget: 2x + 3y ≤ 60 ('at most $60'), (2) Minimum notebooks: x ≥ 10 ('at least 10'), (3) Minimum folders: y ≥ 8 ('at least 8'), (4) Non-negativity: x ≥ 0, y ≥ 0, (5) Integer requirement: x, y ∈ ℤ (whole numbers). Choice A correctly represents all constraints with proper inequality directions and includes the integer requirement. Choice B incorrectly uses 2x + 3y ≥ 60 (spending at least $60 instead of at most), Choice C incorrectly uses x ≤ 10 (at most 10 notebooks instead of at least), and Choice D incorrectly uses equality 2x + 3y = 60 (exactly $60 instead of at most). Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions!
Consider the system of constraints $\begin{cases}2x+y \le 12 \\ x+2y \le 14 \\ x \ge 0 \\ y \ge 0 \end{cases}$ Which point is nonviable (infeasible) for this system?
Answer: $(5,4)$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. Checking viability is systematic: (1) write out each constraint, (2) substitute the proposed solution into each one, (3) verify each is satisfied (inequality holds, equation balances), (4) check context reasonableness (non-negative? integers if needed?). If everything passes, it's viable. If anything fails, it's nonviable—and you should identify WHICH constraint was violated. Complete checking means checking ALL constraints, not just some! Checking each: $(2,4)$ $2(2)+4=8 \le 12$ (✓), $2+2(4)=10 \le 14$ (✓); $(4,2)$ $2(4)+2=10 \le 12$ (✓), $4+2(2)=8 \le 14$ (✓); $(5,4)$ $2(5)+4=14>12$ (✗); $(0,7)$ $0+7=7 \le 12$ (✓), $0+2(7)=14 \le 14$ (✓)—only $(5,4)$ fails. Choice C correctly identifies the nonviable point with complete checking showing it violates $2x + y \le 12$ . Choice A gently, but $(2,4)$ satisfies both inequalities, so it's viable—check substitutions carefully. Constraint identification from context: (1) list every limitation mentioned ('budget $X,$ 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like $x \ge 0$ , $y \ge 0$ (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A small business runs two ad types: online ads and radio ads. Let $x$ = number of online ads and $y$ = number of radio ads. Constraints: - Each online ad costs $40; each radio ad costs $90. - Total ad budget is at most $900. - They want at least 8 total ads. - At most 6 radio ads can be run. - Nonnegativity: $x\ge 0,\ y\ge 0$ . Which system of inequalities represents the constraints?
Answer: $\begin{cases}40x+90y\le 900\\x+y\ge 8\\y\le 6\\x\ge 0,\\ y\ge 0\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! The constraints are cost 40x + 90y ≤ 900 (at most $900), total ads x + y ≥ 8 (at least 8), radio limit y ≤ 6, and nonnegativity x ≥ 0, y ≥ 0. Choice A correctly represents all constraints with the complete system, using the right directions for each. Choice D incorrectly uses y ≥ 6 for radio, but 'at most' is ≤, not requiring more. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A bakery is deciding how many loaves of bread and batches of muffins to make. Let $x$ = loaves of bread and $y$ = batches of muffins. Each loaf uses 0.5 kg of flour and each muffin batch uses 0.8 kg of flour. The bakery has at most 20 kg of flour. The oven time limit is 30 hours; each loaf takes 1 hour and each muffin batch takes 2 hours. They must make at least 10 loaves of bread. Which constraint does the point $(x,y)=(10,12)$ violate, if any? (Assume nonnegative integers.)
Answer: It violates the oven-time constraint. This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. Checking viability is systematic: (1) write out each constraint, (2) substitute the proposed solution into each one, (3) verify each is satisfied (inequality holds, equation balances), (4) check context reasonableness (non-negative? integers if needed?). If everything passes, it's viable. If anything fails, it's nonviable—and you should identify WHICH constraint was violated. Complete checking means checking ALL constraints, not just some! For (10,12): flour 0.5*10 + 0.8*12 = 5 + 9.6 = 14.6 ≤ 20 (satisfied), oven 1*10 + 2*12 = 10 + 24 = 34 > 30 (violated), bread 10 ≥ 10 (satisfied), nonnegative integers (yes)—violates oven time. Choice B correctly identifies the violated constraint with complete substitution. A distractor like Choice D might skip checking oven as 34 ≤ 30, but it's >—ensure you evaluate all inequalities fully! Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A theater is selling adult tickets (x) and student tickets (y). Adult tickets cost $15 and student tickets cost $9. The theater wants revenue of at least $450, but can sell at most 40 tickets total due to seating. Which system of inequalities models the constraints (including non-negativity)?
Answer: $\begin{cases}15x+9y\ge 450\\x+y\le 40\\x\ge 0,\\ y\ge 0\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! Let's identify each constraint: (1) Revenue constraint: 'wants revenue of at least $450' with adult tickets at $15 and student tickets at $9 gives us 15x + 9y ≥ 450; (2) Seating constraint: 'can sell at most 40 tickets total' translates to x + y ≤ 40; (3) Non-negativity: ticket quantities cannot be negative, so x ≥ 0 and y ≥ 0. Choice C correctly represents all constraints with revenue requirement (≥ 450), seating limitation (≤ 40), and non-negativity conditions. Choice A incorrectly uses ≤ for revenue, meaning they want at most $450 instead of at least; Choice B incorrectly uses ≥ for seating, requiring at least 40 tickets instead of limiting to at most 40; Choice D incorrectly uses = for revenue, requiring exactly $450 instead of at least $450. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions!
A student is choosing smoothies. Let $x$ = number of protein smoothies and $y$ = number of fruit smoothies. Constraints: - Protein requirement: at least 30 g total. Protein smoothie has 12 g; fruit smoothie has 6 g. - Calorie limit: at most 500 calories total. Protein smoothie has 220 cal; fruit smoothie has 140 cal. - Nonnegativity: $x\ge 0,\ y\ge 0$ . Determine which constraint the point $(x,y)=(1,4)$ violates (if any).
Answer: It violates the calorie limit. This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. Checking viability is systematic: (1) write out each constraint, (2) substitute the proposed solution into each one, (3) verify each is satisfied (inequality holds, equation balances), (4) check context reasonableness (non-negative? integers if needed?). If everything passes, it's viable. If anything fails, it's nonviable—and you should identify WHICH constraint was violated. Complete checking means checking ALL constraints, not just some! For (1,4), protein: 12(1) + 6(4) = 12 + 24 = 36 ≥ 30 (✓); calories: 220(1) + 140(4) = 220 + 560 = 780 ≤ 500? 780 > 500 (✗); nonnegativity: yes—so violates calories. Choice B correctly identifies the violated constraint with proper reasoning on the calorie check. Choice A gently, but it actually violates calories, as the substitution shows 780 > 500, so not protein. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A student is choosing how many hours to spend tutoring in math and science this week. Let $x$ be hours of math tutoring and $y$ be hours of science tutoring. Constraints: - Total tutoring time is at most 12 hours: $x+y\le 12$ . - The student wants at least twice as much math as science: $x\ge 2y$ . - At least 3 hours of science: $y\ge 3$ . - Non-negativity: $x\ge 0$ , $y\ge 0$ . Determine which solutions are viable: $(6,3)$ , $(5,3)$ , $(8,4)$ .
Answer: $(6,3)$ and $(8,4)$ are viable. This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. Checking viability is systematic: (1) write out each constraint, (2) substitute the proposed solution into each one, (3) verify each is satisfied (inequality holds, equation balances), (4) check context reasonableness (non-negative? integers if needed?). If everything passes, it's viable. If anything fails, it's nonviable—and you should identify WHICH constraint was violated. Complete checking means checking ALL constraints, not just some! For (6,3): 6+3=9≤12, 6≥2*3=6, 3≥3 (all ✓); for (8,4): 8+4=12≤12, 8≥8, 4≥3 (all ✓); but (5,3): 5<6 violates x≥2y. Choice C correctly determines viability with proper reasoning for both points. A distractor like choice A might miss checking (8,4) fully, but gently substitute into each constraint for all points to confirm. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A community garden is deciding how many acres of tomatoes and peppers to plant. Let $x$ be acres of tomatoes and $y$ be acres of peppers. Constraints: - Total land available: at most 12 acres: $x+y\le 12$ . - Water limit: tomatoes need 3 units/acre and peppers need 2 units/acre; at most 30 units: $3x+2y\le 30$ . - The garden must plant at least 2 acres of peppers: $y\ge 2$ . - Non-negativity: $x\ge 0$ , $y\ge 0$ . Which point is viable?
Answer: $(6,4)$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. Checking viability is systematic: (1) write out each constraint, (2) substitute the proposed solution into each one, (3) verify each is satisfied (inequality holds, equation balances), (4) check context reasonableness (non-negative? integers if needed?). If everything passes, it's viable. If anything fails, it's nonviable—and you should identify WHICH constraint was violated. Complete checking means checking ALL constraints, not just some! For (6,4), land: 6+4=10≤12 (✓), water: 3(6)+2(4)=18+8=26≤30 (✓), peppers: 4≥2 (✓), x≥0 y≥0 (✓)—all pass; others like (10,3) fail land 13>12. Choice C correctly determines viability with proper reasoning on the only point satisfying all. A distractor like choice A might overlook land 10+3=13>12, but gently verify each substitution to catch that. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A electronics manufacturer produces smartphones and tablets. Due to component shortages, they can make at most $300$ devices per week. Smartphones require $2$ hours of assembly time each, while tablets require $3$ hours each. The company has $750$ hours of assembly time available weekly. Market demand requires producing at least twice as many smartphones as tablets. Which inequality represents the market demand constraint if $s$ is smartphones and $t$ is tablets?
Answer: $s \geq 2t$ because smartphones must be at least double the tablet quantity When you encounter linear programming problems involving constraints, focus on translating the word problem into mathematical inequalities by identifying the key relationships between variables. The market demand constraint states that the company must produce "at least twice as many smartphones as tablets." This means smartphone production ( $s$ ) must be greater than or equal to two times tablet production ( $t$ ). Mathematically, this translates directly to $s \geq 2t$ . Let's examine why the other options miss the mark: Option A ( $s + t \geq 2t$ ) simplifies to $s \geq t$ , which only requires smartphones to equal or exceed tablets, not be twice as many. This misinterprets the constraint by including total production rather than comparing the individual quantities. Option B ( $2s \geq t$ ) reverses the relationship entirely. This would mean twice the smartphones must exceed tablets, but the problem states smartphones must be at least twice the tablets, not that double the smartphones exceeds tablets. Option D ( $t \leq \frac{s}{2}$ ) is mathematically equivalent to the correct answer when rearranged, but it's written from the tablet perspective rather than directly expressing the smartphone constraint as stated in the problem. Study tip: When translating "at least twice as many of A as B," write it as $A \geq 2B$ . Practice converting word constraints into inequalities by identifying which variable should be isolated and whether the relationship requires "at least" (≥), "at most" (≤), or exact equality (=).
A theater company is planning a production and must consider both budget and space constraints. The cost to produce the show is $8000 plus $15 per ticket printed. The theater has a maximum capacity of $500$ seats, but fire safety regulations require that no more than $480$ people can be present during any performance. If the company needs to keep total production costs under $15000, which system of inequalities correctly models all the constraints on the number of tickets $t$ that can be printed?
Answer: $t \leq 480$ and $8000 + 15t < 15000$ The correct answer is A. The fire safety regulation limits attendance to no more than 480 people, so $t \leq 480$ . The cost constraint requires total costs to be under $15000 (not equal to), so $8000 + 15t < 15000$ . Choice B incorrectly uses the theater capacity (500) instead of the fire safety limit (480). Choice C incorrectly allows costs to equal $15000. Choice D incorrectly uses strict inequality for the attendance constraint when exactly 480 people should be allowed.
A school club is ordering custom T-shirts (x) and custom hoodies (y). Each T-shirt costs $12 and each hoodie costs $28. The club has at most $420 to spend, and they need at least 20 total items. Also, they cannot order negative amounts. Which system of inequalities represents all constraints?
Answer: $\begin{cases}12x+28y\le 420\\x+y\ge 20\\x\ge 0,\\ y\ge 0\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! Let's identify each constraint: (1) Budget constraint: 'at most $420 to spend' with T-shirts at $12 each and hoodies at $28 each gives us 12x + 28y ≤ 420; (2) Quantity constraint: 'need at least 20 total items' translates to x + y ≥ 20; (3) Non-negativity: 'cannot order negative amounts' means x ≥ 0 and y ≥ 0. Choice B correctly represents all constraints with the complete system including budget limitation (≤), minimum quantity requirement (≥), and non-negativity conditions. Choice A incorrectly uses ≥ for the budget constraint, which would mean they must spend at least $420 rather than at most; Choice C incorrectly uses ≤ for the quantity constraint, limiting them to at most 20 items instead of requiring at least 20; Choice D omits the crucial non-negativity constraints. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions!
A caterer is choosing trays of sandwiches and trays of fruit for an event. Let $x =$ number of sandwich trays and $y =$ number of fruit trays. Each sandwich tray serves 8 people and costs $\24$ . Each fruit tray serves 6 people and costs $\18$ . The caterer must serve at least 60 people and spend no more than $\180$ . Trays must be whole numbers and nonnegative. Which constraint system matches the situation?
Answer: $\begin{cases}24x+18y \le 180\\8x+6y \ge 60\\x \ge 0,\\ y \ge 0\\x,y \in \mathbb{Z}\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost $\le 500$ , 'need at least 10 units' becomes quantity $\ge 10$ . The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! The constraints are cost $24x + 18y \le 180$ (no more than $180), serving $8x + 6y \ge 60$ (at least 60 people), and $x \ge 0$ , $y \ge 0$ with integers (nonnegative whole trays). Choice A correctly represents all constraints with $\le$ for cost and $\ge$ for serving, including nonnegativity and integers. A distractor like Choice B flips cost to $\ge 180$ , which means spending at least $180, but the limit is at most—remember to match 'no more than' to $\le$ ! Constraint identification from context: (1) list every limitation mentioned ('budget $X,$ 'time $\le$ Y hours,' 'need $\ge$ Z units'), (2) translate using key phrases: 'at most' $\to \le$ , 'at least' $\to \ge$ , 'exactly' $\to =$ , 'more than' $\to >$ , 'less than' $\to <$ , (3) don't forget implicit constraints like $x \ge 0$ , $y \ge 0$ (can't be negative) or $x, y$ integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A school club is ordering snacks for a movie night. Let $x$ = number of pizzas and $y$ = number of boxes of cookies. Constraints: - Each pizza costs $12 and each cookie box costs $5. - The club can spend at most $120. - They need at least 15 total items (pizzas + cookie boxes). - They cannot order negative amounts. Which system of inequalities represents all constraints?
Answer: $\begin{cases}12x+5y\le 120\\x+y\ge 15\\x\ge 0,\\ y\ge 0\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! The constraints are cost 12x + 5y ≤ 120 (at most $120), total items x + y ≥ 15 (at least 15), and nonnegativity x ≥ 0, y ≥ 0, forming the complete system. Choice B correctly represents all constraints with the complete system, including the correct direction for the budget inequality. Choice A incorrectly uses ≥ for the cost, but 'at most' translates to ≤, which would allow overspending. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A theater is selling two types of tickets: adult and student. Let $a$ = number of adult tickets and $s$ = number of student tickets. Constraints: - The theater can seat at most 180 people: $a+s\le 180$ . - At least 40 student tickets must be sold: $s\ge 40$ . - Revenue goal: adult tickets are $15 and student tickets are $10; total revenue must be at least $2200. - Nonnegativity: $a\ge 0,\ s\ge 0$ . Which of the following points is viable?
Answer: $(a,s)=(120,50) This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. Checking viability is systematic: (1) write out each constraint, (2) substitute the proposed solution into each one, (3) verify each is satisfied (inequality holds, equation balances), (4) check context reasonableness (non-negative? integers if needed?). If everything passes, it's viable. If anything fails, it's nonviable—and you should identify WHICH constraint was violated. Complete checking means checking ALL constraints, not just some! Checking each: (120,50) seats 170 ≤ 180 (✓), s=50 ≥ 40 (✓), revenue 15(120)+10(50)=1800+500=2300 ≥ 2200 (✓); (100,30) s=30 < 40 (✗); (160,40) seats 200 > 180 (✗); (150,35) seats 185 > 180 (✗) and s=35 < 40 (✗)—only (120,50) works. Choice A correctly determines viability with complete system checking for all points. Choice B gently, but (100,30) violates minimum students, though others fail differently—always check all. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A community garden is planting tomatoes (x) and peppers (y). The garden has at most 30 plots total, so $x+y\le 30$ . Tomatoes require 3 units of compost per plot and peppers require 1 unit per plot; at most 60 compost units are available, so $3x+y\le 60$ . Also, at least 8 plots must be peppers: $y\ge 8$ . Non-negativity applies. Which constraint does the plan $(x,y)=(18,10)$ violate, if any?
Answer: It violates $3x+y\le 60$ . This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. Checking viability is systematic: (1) write out each constraint, (2) substitute the proposed solution into each one, (3) verify each is satisfied (inequality holds, equation balances), (4) check context reasonableness (non-negative? integers if needed?). If everything passes, it's viable. If anything fails, it's nonviable—and you should identify WHICH constraint was violated. Complete checking means checking ALL constraints, not just some! Let's check (18, 10) against each constraint: (1) Total plots: x + y ≤ 30 → 18 + 10 = 28 ≤ 30 ✓; (2) Compost units: 3x + y ≤ 60 → 3(18) + 10 = 54 + 10 = 64 > 60 ✗ VIOLATED!; (3) Pepper minimum: y ≥ 8 → 10 ≥ 8 ✓; (4) Non-negativity: 18 ≥ 0 ✓ and 10 ≥ 0 ✓. The plan violates the compost constraint! Choice B correctly identifies that the plan violates 3x + y ≤ 60 because 3(18) + 10 = 64 > 60, making the plan nonviable. Choice A incorrectly claims it violates x + y ≤ 30, but 28 ≤ 30 is satisfied; Choice C incorrectly claims it violates y ≥ 8, but 10 ≥ 8 is satisfied; Choice D incorrectly claims the plan is viable when it clearly violates the compost constraint. The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
An event planner is ordering chairs and tables for a fundraiser. Let $x$ be the number of chairs and $y$ be the number of tables. Constraints: - Each chair costs $8 and each table costs $25; budget at most $400: $8x+25y\le 400$ . - Seating requirement: each table requires 4 chairs, so there must be at least 4 chairs per table: $x\ge 4y$ . - At least 6 tables are needed: $y\ge 6$ . - Chairs and tables must be whole numbers and nonnegative. Which constraint does $(x,y)=(20,7)$ violate?
Answer: Seating requirement $x\ge 4y$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. Checking viability is systematic: (1) write out each constraint, (2) substitute the proposed solution into each one, (3) verify each is satisfied (inequality holds, equation balances), (4) check context reasonableness (non-negative? integers if needed?). If everything passes, it's viable. If anything fails, it's nonviable—and you should identify WHICH constraint was violated. Complete checking means checking ALL constraints, not just some! For (20,7), budget: 8(20)+25(7)=160+175=335≤400 (✓), seating: 20≥4*7=28? 20<28 (✗), tables: 7≥6 (✓), integers nonneg (✓)—violates seating. Choice B correctly identifies the violated constraint with complete checking. A distractor like choice A might ignore seating and only check budget, but gently remember to test all constraints systematically. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A school club is making gift bags using stickers and pencils. Let $x$ be the number of sticker packs and $y$ be the number of pencil packs purchased. Each sticker pack costs $4 and each pencil pack costs $6. The club has at most $72 to spend, needs at least 10 total packs, and cannot buy negative amounts. Which system of inequalities represents all constraints?
Answer: $\begin{cases}4x+6y\le 72\\ x+y\ge 10\\ x\ge 0,\\ y\ge 0\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! The constraints here are cost 4x + 6y ≤ 72 from the $72 budget, total packs x + y ≥ 10 for at least 10 packs, and non-negativity x ≥ 0, y ≥ 0 since you can't buy negative amounts. Choice B correctly represents all constraints with the complete system, including the minimum total packs as ≥ instead of ≤. A common distractor like choice A flips the total packs to ≤ 10, but remember 'at least' means ≥, so gently correct by double-checking the direction of inequalities from key phrases like 'at most' or 'at least.' Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!
A school club is ordering T-shirts and hoodies for a fundraiser. Let $x$ = number of T-shirts and $y$ = number of hoodies. Constraints: - Each T-shirt costs $8 and each hoodie costs $20; the club has at most $400 to spend. - They need at least 30 total items. - They must order at least 10 hoodies. - They cannot order negative amounts. Which system of inequalities represents all constraints?
Answer: $\begin{cases}8x+20y\le 400\\x+y\ge 30\\y\ge 10\\x\ge 0,\\ y\ge 0\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! Let's translate each constraint: (1) Budget constraint: 8x + 20y ≤ 400 (at most $400), (2) Total items: x + y ≥ 30 (at least 30), (3) Minimum hoodies: y ≥ 10 (at least 10), (4) Non-negativity: x ≥ 0, y ≥ 0. Choice A correctly represents all constraints with the proper inequality directions matching the problem's language. Choice B incorrectly uses 8x + 20y ≥ 400, which would mean spending at least $400 rather than at most $400. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions!
A rectangular dog park will have length $L$ and width $W$ (in meters). The fencing budget allows a perimeter of at most 60 meters, and the park must have area at least 200 square meters. Also $L>0$ and $W>0$ . Which system correctly represents these constraints?
Answer: $\begin{cases}2L+2W\le 60\\LW\ge 200\\L>0,\\ W>0\end{cases}$ This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most $500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! Let's identify each constraint: (1) Perimeter constraint: 'perimeter of at most 60 meters' for a rectangle with perimeter = 2L + 2W gives us 2L + 2W ≤ 60; (2) Area constraint: 'area at least 200 square meters' with area = L × W gives us LW ≥ 200; (3) Positivity: dimensions must be positive (not just non-negative), so L > 0 and W > 0. Choice A correctly represents all constraints with perimeter limitation (2L + 2W ≤ 60), minimum area requirement (LW ≥ 200), and strict positivity conditions. Choice B incorrectly uses ≥ for perimeter, requiring at least 60 meters instead of limiting to at most 60; Choice C incorrectly uses L + W for perimeter (missing the factor of 2) and ≤ for area (limiting instead of requiring minimum); Choice D incorrectly represents area as L + W instead of LW. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions!

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