Solve Real-World System Problems
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8th Grade Math › Solve Real-World System Problems
A class is buying notebooks and pens. Notebooks cost $3 each and pens cost $1 each. The class buys 24 items total and spends $52. Let $n$ be the number of notebooks and $p$ be the number of pens. Which system correctly represents the situation?
$n+p=24$ and $3n+p=52$
$n+p=52$ and $3n+p=24$
$3n+p=24$ and $n-p=52$
$n+p=24$ and $n+3p=52$
Explanation
This problem tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns, along with interpreting the solution in context. The process involves: (1) defining variables with units (let n = number of notebooks, p = number of pens), (2) writing the total items (n + p = 24), (3) writing the total cost (3n + p = 52), (4) solving by elimination (subtract: 2n = 28, n = 14, p = 10), (5) interpreting (14 notebooks, 10 pens), and (6) verifying. The correct system is n + p = 24 and 3n + p = 52, as in choice B. These equations are correct because they match the item count and costs given. Common errors include swapping totals or costs, like using n + p = 52 for items. To set up correctly: identify costs and totals, form equations accordingly, solve reliably, interpret purchases, and verify. Avoid errors like misassigning coefficients or not checking sensibility.
A rectangle has perimeter 40 cm. Its length is 3 cm more than its width. Let $l$ be the length and $w$ be the width. Solve the system to find the dimensions of the rectangle.
$w=9.5$ cm and $l=12.5$ cm
$w=10$ cm and $l=13$ cm
$w=7$ cm and $l=13$ cm
$w=8.5$ cm and $l=11.5$ cm
Explanation
This problem tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns, along with interpreting the solution in context. The process involves: (1) defining variables with units (let l = length in cm, w = width in cm), (2) writing the perimeter equation (2l + 2w = 40 or l + w = 20), (3) writing the relation (l = w + 3), (4) solving by substitution (w + 3 + w = 20, 2w = 17, w = 8.5, l = 11.5), (5) interpreting (width 8.5 cm, length 11.5 cm), and (6) verifying perimeter and relation. The correct system is l + w = 20 and l = w + 3, yielding w = 8.5 cm and l = 11.5 cm. These equations are correct as they simplify the perimeter and capture the dimension difference. Common errors include forgetting to halve the perimeter or reversing the relation, leading to integer but incorrect dimensions like w=7, l=13. To set up correctly: express perimeter properly, define relations clearly, solve accurately, ensure dimensions make sense, and verify. Avoid errors like using incorrect coefficients or not checking units.
Two friends’ ages add to 45. One friend is 5 years older than the other. Let $x$ be the older friend’s age and $y$ be the younger friend’s age. Write a system of equations and solve for $x$ and $y$.
$x=25,\ y=20$
$x=30,\ y=15$
$x=20,\ y=25$
$x=22,\ y=23$
Explanation
This problem tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns, along with interpreting the solution in context. The process involves: (1) defining variables with units (let x = older friend's age in years, y = younger friend's age in years), (2) writing the equation from the sum constraint (x + y = 45), (3) writing the equation from the difference constraint (x = y + 5), (4) solving the system using substitution (substitute into first: y + 5 + y = 45, 2y = 40, y = 20, x = 25), (5) interpreting the results (older friend is 25 years old, younger is 20), and (6) verifying both conditions. The correct system is x + y = 45 and x = y + 5, yielding x = 25 and y = 20. These equations are correct because they capture the total age and the age difference accurately. Common errors include reversing the difference equation or arithmetic slips leading to pairs like x=20, y=25, which don't make sense contextually. To set up correctly: identify the relationships carefully, define variables with units, form equations directly from statements, solve step-by-step, interpret sensibly (ages positive and logical), and verify. Avoid errors like assuming the wrong variable is older or not checking if the solution fits the real-world context.
Two friends’ ages add to 45. One friend is 5 years older than the other. Let $x$ be the older friend’s age and $y$ be the younger friend’s age. Write a system of equations and solve for $x$ and $y$.
$x=22,\ y=23$
$x=20,\ y=25$
$x=25,\ y=20$
$x=30,\ y=15$
Explanation
This problem tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns, along with interpreting the solution in context. The process involves: (1) defining variables with units (let x = older friend's age in years, y = younger friend's age in years), (2) writing the equation from the sum constraint (x + y = 45), (3) writing the equation from the difference constraint (x = y + 5), (4) solving the system using substitution (substitute into first: y + 5 + y = 45, 2y = 40, y = 20, x = 25), (5) interpreting the results (older friend is 25 years old, younger is 20), and (6) verifying both conditions. The correct system is x + y = 45 and x = y + 5, yielding x = 25 and y = 20. These equations are correct because they capture the total age and the age difference accurately. Common errors include reversing the difference equation or arithmetic slips leading to pairs like x=20, y=25, which don't make sense contextually. To set up correctly: identify the relationships carefully, define variables with units, form equations directly from statements, solve step-by-step, interpret sensibly (ages positive and logical), and verify. Avoid errors like assuming the wrong variable is older or not checking if the solution fits the real-world context.
A student solves a system for a fundraiser and gets $(x,y)=(18,12)$, where $x$ is the number of large water bottles and $y$ is the number of small water bottles. Large bottles cost $2 each and small bottles cost $1 each. The student also knows 30 bottles were sold total. What does $y=12$ mean in this context?
$12 was collected in total.
Each bottle cost $12.
12 small bottles were sold.
12 large bottles were sold.
Explanation
This problem tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns, along with interpreting the solution in context. The process involves: (1) defining variables with units (let x = large bottles, y = small bottles), (2) writing total bottles (x + y = 30), (3) writing revenue if needed (but here solution given), (4) interpreting the given solution (x=18 large, y=12 small), (5) verifying it fits (18+12=30), and (6) explaining y=12 as 12 small bottles sold. The value y=12 means 12 small bottles were sold. This interpretation is correct as it directly ties to the variable definition and context. Common errors include misinterpreting variables or ignoring the total. To set up correctly: understand given solution, relate to context, verify consistency, and explain clearly. Avoid errors like confusing x and y or adding extraneous details.
A class is buying notebooks and pens. Notebooks cost $3 each and pens cost $1 each. The class buys 24 items total and spends $52. Let $n$ be the number of notebooks and $p$ be the number of pens. Which system correctly represents the situation?
$n+p=24$ and $3n+p=52$
$n+p=52$ and $3n+p=24$
$3n+p=24$ and $n-p=52$
$n+p=24$ and $n+3p=52$
Explanation
This problem tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns, along with interpreting the solution in context. The process involves: (1) defining variables with units (let n = number of notebooks, p = number of pens), (2) writing the total items (n + p = 24), (3) writing the total cost (3n + p = 52), (4) solving by elimination (subtract: 2n = 28, n = 14, p = 10), (5) interpreting (14 notebooks, 10 pens), and (6) verifying. The correct system is n + p = 24 and 3n + p = 52, as in choice B. These equations are correct because they match the item count and costs given. Common errors include swapping totals or costs, like using n + p = 52 for items. To set up correctly: identify costs and totals, form equations accordingly, solve reliably, interpret purchases, and verify. Avoid errors like misassigning coefficients or not checking sensibility.
A rectangle has perimeter 40 cm. Its length is 3 cm more than its width. Let $l$ be the length and $w$ be the width. Solve the system to find the dimensions of the rectangle.
$w=7$ cm and $l=13$ cm
$w=9.5$ cm and $l=12.5$ cm
$w=10$ cm and $l=13$ cm
$w=8.5$ cm and $l=11.5$ cm
Explanation
This problem tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns, along with interpreting the solution in context. The process involves: (1) defining variables with units (let l = length in cm, w = width in cm), (2) writing the perimeter equation (2l + 2w = 40 or l + w = 20), (3) writing the relation (l = w + 3), (4) solving by substitution (w + 3 + w = 20, 2w = 17, w = 8.5, l = 11.5), (5) interpreting (width 8.5 cm, length 11.5 cm), and (6) verifying perimeter and relation. The correct system is l + w = 20 and l = w + 3, yielding w = 8.5 cm and l = 11.5 cm. These equations are correct as they simplify the perimeter and capture the dimension difference. Common errors include forgetting to halve the perimeter or reversing the relation, leading to integer but incorrect dimensions like w=7, l=13. To set up correctly: express perimeter properly, define relations clearly, solve accurately, ensure dimensions make sense, and verify. Avoid errors like using incorrect coefficients or not checking units.
Two cars start 300 miles apart and drive toward each other. Car A drives 60 mph and Car B drives 40 mph. Let $t$ be the time in hours until they meet, and let $d$ be the distance Car A travels. Which system correctly models the situation and gives the time they meet?
System: $d=60t$ and $d+40t=300$; Solution: $t=3$ hours
System: $d=40t$ and $d+60t=300$; Solution: $t=2$ hours
System: $d=60t$ and $d-40t=300$; Solution: $t=5$ hours
System: $d=60t$ and $d+40t=300$; Solution: $t=5$ hours
Explanation
This problem tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns, along with interpreting the solution in context. The process involves: (1) defining variables with units (let t = time in hours, d = distance Car A travels in miles), (2) writing the equation for Car A's distance (d = 60t), (3) writing the equation for total distance (d + 40t = 300), (4) solving by substitution (60t + 40t = 300, 100t = 300, t = 3), (5) interpreting (they meet after 3 hours), and (6) verifying distances add to 300. The correct system is d = 60t and d + 40t = 300, yielding t = 3 hours. These equations are correct because they account for each car's speed and the closing distance. Common errors include switching speeds or using subtraction instead of addition, leading to incorrect times like t=5. To set up correctly: define variables based on the problem's request, model directions accurately, solve consistently, interpret in context, and verify. Avoid errors like miscalculating the solution or ignoring relative motion.
A rectangle has perimeter 40 cm. Its length is 3 cm more than its width. Let $\ell$ be the length and $w$ be the width. What are the dimensions of the rectangle?
$w=7\text{ cm},\ \ell=10\text{ cm}$
$w=9\text{ cm},\ \ell=12\text{ cm}$
$w=8.5\text{ cm},\ \ell=11.5\text{ cm}$
$w=8\text{ cm},\ \ell=11\text{ cm}$
Explanation
This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as w for width and ℓ for length in cm, (2) writing the perimeter equation: 2w + 2ℓ = 40 or w + ℓ = 20, (3) writing ℓ = w + 3, (4) solving by substitution, (5) interpreting, and (6) verifying. Solving correctly: (w + 3) + w = 20, 2w + 3 = 20, 2w = 17, w = 8.5, ℓ = 11.5. These satisfy: perimeter 2(8.5 + 11.5) = 40 and difference 3. Common errors include forgetting to double sides or misinterpreting 'more than.' To avoid errors: recall perimeter formula, define variables with units, translate relations accurately, solve, accept decimals if sensible, and verify.
A store sells notebooks and pens. A notebook costs $3 and a pen costs $1. A student buys 18 items total and spends $38. Let $n$ be the number of notebooks and $p$ be the number of pens. How many notebooks did the student buy?
8
10
12
14
Explanation
This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as n for notebooks and p for pens, (2) writing n + p = 18 for total items, (3) writing 3n + p = 38 for total cost, (4) solving, (5) interpreting (number of notebooks), and (6) verifying. Solving correctly: subtract first from second: (3n + p) - (n + p) = 38 - 18, 2n = 20, n = 10, then p = 8. These satisfy: 10 + 8 = 18 and 3(10) + 8 = 30 + 8 = 38. Common errors include incorrect cost equation or not solving the system fully. To avoid errors: define clearly, translate costs and counts, use elimination or substitution, interpret the asked quantity, and verify.