5. Modeling with Systems of Linear Equations
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Chapter 5
5.
Modeling with Systems of Linear Equations
In the content, systems of equations are portrayed as invaluable tools for solving real-world problems. For example, if you're running a car dealership with your father, you can use these equations to determine the number of cars and motorcycles on the lot based on the total number of tires. Similarly, when planning a river safari in Sri Lanka, systems of equations can help figure out how many adults and children can safely be on a boat, considering weight restrictions and the amount of gear. These practical applications make systems of equations not just a mathematical concept but a life skill.
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Lesson Settings & Tools
| | 7 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Modeling with Systems of Linear Equations
Slide of 7
In this lesson, systems of equations will be used to model and solve real-life situations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
How Many Cars and Motorcycles?
Mark has just started working with his father at a car dealership. They sell cars and motorcycles.
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Example
Extra Dollar
Zain volunteered to work at the reception desk at the school concert.
At the end of the evening, they picked up the cash box and noticed a dollar lying on the floor next to it. Zain wonders whether the dollar belongs inside the cash box or not.
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Hint
Without considering the extra dollar found on the floor, write and solve a system of equations. Does the solution make sense in this context?
Solution
To find whether the dollar belongs inside the cash box, a system of equations will be written and solved. Let x and y be the number of tickets sold to individuals and to couples, respectively. Considering that 47 tickets were sold, an equation can be written.
It has been found that x=12. Now, this value will be substituted into the first equation to find the value of y.
The solution to the system is x=12 and y=35. Considering the context, this means that without the extra dollar, 35 tickets were sold to couples and 12 tickets were sold to individuals. However, this is not possible, since couples tickets were only sold in pairs. Therefore, the dollar found on the floor should be placed inside the cash box.
x+y=47
Next, since each individual's ticket is $5, the total amount of money made by selling tickets to individuals is 5x. Similarly, since each ticket sold to couples is $4, the total amount of money made by selling tickets to couples is 4y. Considering that the cash box contains $200 total, a second equation can be written.
5x+4y=200
These two equations can be written together to form a system of equations.
{x+y=475x+4y=200(I)(II)
Finally, the above system will be solved. For simplicity, the Substitution Method will be used.
{x+y=475x+4y=200
SubEqn
(I): LHS−x=RHS−x
{y=47−x5x+4y=200
Substitute
(II): y=47−x
{y=47−x5x+4(47−x)=200
{y=47−xx=12
Extra
The answer can also be found by solving a system considering the extra dollar.
This means that, considering the extra dollar, 34 tickets were sold to couples and 13 tickets were sold to individuals. In the context of the situation, this is a perfectly valid answer.
{x+y=475x+4y=201(I)(II)
The Substitution Method will be used.
{x+y=475x+4y=201
▼
Solve by substitution
SubEqn
(I): LHS−x=RHS−x
{y=47−x5x+4y=201
Substitute
(II): y=47−x
{y=47−x5x+4(47−x)=201
Distr
(II): Distribute 4
{y=47−x5x+188−4x=201
SubTerm
(II): Subtract term
{y=47−xx+188=201
SubEqn
(II): LHS−188=RHS−188
{y=47−xx=13
Substitute
(I): x=13
{y=47−13x=13
SubTerm
(I): Subtract term
{y=34x=13
Example
Old and New Pennies
Pennies used to be made of 95% copper. In 1983, the cost of this metal significantly increased, so the US government changed the composition of the coin. Since 1983, pennies have been made of 97.5% zinc and plated with a thin copper coating.
Pennies made in 1983 and after weigh 2.5 grams, while pennies made before 1983 weigh 3.11 grams. Knowing that a roll of pennies contains 50 coins and weighs 138.42 grams, how many of the older, heavier pennies and newer, lighter pennies does this roll contain?
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Hint
Write and solve a system of equations.
Solution
To find out how many of the older, heavier pennies and how many of the newer, lighter pennies this roll contains, a system of equations will be written and solved. Let x be the number of older pennies and y the number of newer pennies in the roll. Since the roll contains 50 coins, an equation can be written.
Now, Equation (I) will be subtracted from Equation (II).
It has been found that x=22. Finally, this value will be substituted in the first equation to find the value y.
The solution to the system is x=22 and y=28. In the context of the situation, this means that in the 50-coin roll there are 22 older, heavier pennies and 28 newer, lighter pennies.
x+y=50
Next, since older pennies weigh 3.11 grams, the total weight of the older pennies in the roll is 3.11x. Similarly, since a newer penny weighs 2.5 grams, the total weight of the newer pennies is 2.5y. With this information, and knowing that the 50-coin roll weighs 138.42 grams, a second equation can be written.
3.11x+2.5y=138.42
These two linear equations form a system of linear equations.
{x+y=503.11x+2.5y=138.42(I)(II)
For simplicity the system will be solved by the Elimination Method. To do so, the first equation will be multiplied by 2.5 in order for the y-variable to have the same coefficient in both equations. {x+y=503.11x+2.5y=138.42
MultEqn
(I): LHS⋅2.5=RHS⋅2.5
{(x+y)(2.5)=50(2.5)3.11x+2.5y=138.42
{2.5x+2.5y=1253.11x+2.5y=138.42
{2.5x+2.5y=1253.11x+2.5y=138.42
SysEqnSub
(II): Subtract (I)
{2.5x+2.5y=1253.11x+2.5y−(2.5x+2.5y)=138.42−125
{2.5x+2.5y=125x=22
Example
Orange and Peach Juice
Tiffaniqua is selling juice to make some money for a trip to the beach. To prepare a big jug, she used oranges and peaches.
Tiffaniqua's math teacher stopped by to buy some juice and told her the amount of carbohydrates that oranges and peaches have.
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Hint
Write and solve a system of equations.
Solution
To find out how many oranges and peaches Tiffaniqua used for a jug, a system of equations will be written and solved. Let x and y be the number of oranges and peaches, respectively. A first equation can be written knowing that Tiffaniqua used 13 fruits.
Now that both equations are written in slope-intercept form, their slopes and y-intercepts can be used to draw the lines on the same coordinate plane. 
x+y=13
Next, consider the fact that the jug contains 128 grams of carbohydrates. It is also known that an average orange has about 12 grams of carbohydrates and that an average peach has about 8 grams of carbohydrates. With all this information, a second equation can be written.
12x+8y=128
These two equations form a system of linear equations.
{x+y=1312x+8y=128(I)(II)
For simplicity, the system will be solved by graphing. To do so, both linear equations must first be written in slope-intercept form.
{x+y=1312x+8y=128
▼
(I), (II): Write in slope-intercept form
SubEqn
(I): LHS−x=RHS−x
{y=-x+1312x+8y=128
SubEqn
(II): LHS−12x=RHS−12x
{y=-x+138y=-12x+128
DivEqn
(II): LHS/8=RHS/8
{y=-x+13y=8-12x+128
WriteSumFrac
(II): Write as a sum of fractions
{y=-x+13y=8-12x+8128
MovePartNumRight
(II): ca⋅b=ca⋅b
{y=-x+13y=8-12x+8128
MoveNegNumToFrac
(II): Put minus sign in front of fraction
{y=-x+13y=-812x+8128
ReduceFrac
(II): ba=b/4a/4
{y=-x+13y=-23x+8128
CalcQuot
(II): Calculate quotient
{y=-x+13y=-23x+16
Finally, the point of intersection can be determined.
The lines intersect at the point with coordinates (6,7). Therefore, the solution to the system is x=6 and y=7. In the context of the situation, this means that Tiffaniqua used 6 oranges and 7 peaches to prepare a jug of juice.
Example
Adventure on the Water
Adventures on the Water is a company that organizes river safaris in Sri Lanka. They take tourists on boats along a river in the middle of the jungle for a full day.
Each boat can hold at most 8 people and can only carry 1200 pounds of weight, including passengers and gear, for safety reasons. The company assumes that, on average, an adult weighs 150 pounds and a child weighs 75 pounds. It is also assumed that each group will require 200 pounds of gear plus 10 pounds of gear per person. There are three groups who wish to take a river safari.
- Group A has 4 adults and 2 children.
- Group B has 3 adults and 5 children.
- Group C has 8 adults.
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Hint
Write and solve a system of inequalities.
Solution
To determine which of the groups can safely take a safari, a system of linear inequalities will be written and solved. Let x be the number of adults and y the number of children. Knowing that each boat can hold at most 8 people, the first inequality can be written.
Now both lines can be drawn on the same coordinate plane. Since the number of adults and children cannot be negative, only the first quadrant will be considered. Also, because neither inequality is strict, the lines will be solid. 
Next, a point not on the boundary lines will be tested to determine the region to be shaded. For simplicity, (1,2) will be used. To start, this point will be tested in the first inequality. If a true statement is obtained, the region containing the point will be shaded. Otherwise, the opposite region will be shaded.
Since a true statement was obtained, the region that includes this point will be shaded. 
x+y≤8(I)
Next, a second inequality will be written. The company assumes that an adult weighs 150 and a child weighs 75 pounds. Also, for each person there are considered 10 extra pounds of gear.
Adult: Child: 150x+10x=160x175y+10y=85y
Furthermore, each group requires 200 pounds of gear, no matter the number of people. Knowing that each boat can only carry 1200, the second inequality can be written.
160x+85y+200≤1200(II)
The two inequalities obtained form a system of inequalities.
{x+y≤8160x+85y+200≤1200(I)(II)
The above system will be solved by graphing. To do so, the equations of the boundary lines will be written in slope-intercept form. To obtain the equations of the boundaries lines, the inequality symbols will be replaced with equals signs.
{x+y=8160x+85y+200=1200
▼
(I), (II): Write in slope-intercept form
SubEqn
(I): LHS−x=RHS−x
{y=-x+8160x+85y+200=1200
SubEqn
(II): LHS−200=RHS−200
{y=-x+8160x+85y=1000
SubEqn
(II): LHS−160x=RHS−160x
{y=-x+885y=-160x+1000
DivEqn
(II): LHS/85=RHS/85
{y=-x+8y=85-160x+1000
WriteSumFrac
(II): Write as a sum of fractions
{y=-x+8y=85-160x+851000
MoveNegNumToFrac
(II): Put minus sign in front of fraction
{y=-x+8y=-85160x+851000
MovePartNumRight
(II): ca⋅b=ca⋅b
{y=-x+8y=-85160x+851000
ReduceFrac
(II): ba=b/5a/5
{y=-x+8y=-1732x+17200
By following the same procedure, the region that corresponds to the second inequality can be determined.
| Test Point: (1,2) | ||
|---|---|---|
| Inequality | Substitute | Simplify |
| x+y≤8 | 1+2≤?8 | 3≤8 ✓ |
| 160x+85y+200≤1200 | 160(1)+85(2)+200≤?1200 | 230≤1200 ✓ |
The second inequality is also satisfied by (1,2). Therefore, the region that contains this point will be shaded.
To fully see the region that satisfies both inequalities, the unwanted regiones will be removed.
Finally, a point that represents each of the three groups will be plotted to see if they belong to the shaded area.
| Group | Adults & Children | Point |
|---|---|---|
| A | 4 adults and 2 children | (4,2) |
| B | 3 adults and 5 children | (3,5) |
| C | 8 adults | (8,0) |
These points will be plotted on the coordinate plane.
The points that represent groups A and B are in the shaded area, and the point that represents group C is not in the shaded area. Therefore, A and B are the only groups that can safely take a river safari.
Closure
How Many Cars and Motorcycles?
The challenge presented at the beginning of the lesson can also be modeled by a system of equations. It is known that Mark and his father sell cars and motorcycles.
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Hint
Write and solve a system of equations.
Solution
To find the number of a cars and motorcycles in the agency, a system of linear equations will be written and solved. Let x be the number of cars and y the number of motorcycles. Because there are 18 vehicles, the sum of the two variables equals 18.
The solution to the system is x=7 and y=11. In this context, it means that there are 7 cars and 11 motorcycles in the agency.
x+y=18
Next, since spare tires are not considered, each car has 4 tires and each motorcycle has 2 tires. Knowing that Mark counted 50 tires, a second equation can be written.
4x+2y=50
These two linear equations form a system.
{x+y=184x+2y=40(I)(II)
Finally, the system will be solved. For simplicity, the Substitution Method will be used.
{x+y=184x+2y=50
▼
(I), (II): Solve by substitution
SubEqn
(I): LHS−x=RHS−x
{y=-x+184x+2y=50
Substitute
(II): y=-x+18
{y=-x+184x+2(-x+18)=50
Distr
(II): Distribute 2
{y=-x+184x−2x+36=50
SubTerm
(II): Subtract term
{y=-x+182x+36=50
SubEqn
(II): LHS−36=RHS−36
{y=-x+182x=14
DivEqn
(II): LHS/2=RHS/2
{y=-x+18x=7
Substitute
(I): x=7
{y=-7+18x=7
AddTerms
(I): Add terms
{y=11x=7
Modeling with Systems of Linear Equations
Exercise 1.1
A metal alloy is the mixture of 2 or more types of metal. A jewelry supplier sells two metal alloys. One alloy is 40% silver and the other is 20% silver. How much of each alloy must be combined to create 5 kilograms of an alloy containing 25% silver?
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Let x represent the weight of the 40 % silver alloy in kilograms and y represent the weight of the 20 % silver alloy in kilograms. The sum of these two alloys should be equal to 5 kilograms. x+y=5 To create a second equation, we will first rewrite percentages as decimals. ccc Percent & → & Decimal [0.5em] 40 % &→ & 0.4 20 % &→ & 0.2 25 % &→ & 0.25 We want 5 kilograms of an alloy that contains 25 % silver. With this information in mind and using the above percentages, we can write the following equation. 0.4x+ 0.2y= 0.25(5) ⇕ 0.4x+0.2y=1.25 We can form our system of equations by combining the two equations. x+y=5 & (I) 0.4x+0.2y=1.25 & (II) Let's solve the system by using the Substitution Method. We will first isolate y in the first equation and substitute it into the second equation to find the value of x.
Next, we can substitute 54 for x in Equation (I), and solve for y.
Therefore, we need 154 kilograms of the alloy that is 40 % silver and 54 kilograms of the alloy that is 20 % silver. These values can also be expressed as decimal numbers. ccc Fraction & &Decimal [1em] 15/4 & → &3.75 [1em] 5/4 & → &1.25
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A scooter store costs $2000 per month to operate. The store pays an average of $50 per scooter from the manufacturer. The average selling price of each scooter at the store is $100. How many scooters must the store sell each month in order to make it to the break even point?
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To find the break even point for the scooter store, we will write a system of equations. Let s be the number of scooters sold and d be the amount of money from the sale of scooters. The store sells scooters for $100 each, so the income d can be represented by the following equation. d=100s Buying a scooter from the manufacturer costs the store $50. Additionally, the cost of operating the store each month is $2000. The expenses the store incurs each month is equal to the sum of these two costs. Since we are asked to find the break even point for the store, the sum must be equal to d. d=50s+2000 With these two equations, we can form a system of equations. d=100s & (I) d=50s+2000 & (II) If we solve this system of equations, we can find the monthly break even point for the store. To determine the number of scooters the store needs to sell, we solve for s. Let's use the Substitution Method.
The break even point is reached when 40 scooters are sold. Therefore, the store needs to sell at least 40 scooters per month to avoid loss.
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Vincenzo has a jar of quarters and pennies. He wants to choose 12 coins that are worth exactly $2.28. How many quarters and pennies does Vincenzo have to take from the jar?
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Let p represent the number of pennies and q the number of quarters. We can write an equation knowing that Vincenzo wants to choose 12 coins. p+q=12 Vincenzo wants the coins to have a value of $ 2.28. Knowing that the value of a penny is $ 0.01 and that the value of a quarter is $ 0.25, we can write a second equation. 0.01p+ 0.25q= 2.28 Then we can combine these two equations we can form a system of equations. p+q=12 & (I) 0.01p+0.25q=2.28 & (II) Let's solve this system by using the Substitution Method. We will start by isolating p in Equation (I). Then, we will substitute its equivalent expression into Equation (II).
Now that we know the value of q, we can substitute it into Equation (I) to find the value of p.
Therefore, Vincenzo has to take 3 pennies and 9 quarters in order to make $ 2.28 from the 12 coins.
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The Panthers football team scored 68 points during the game last night. They made one more field goal than twice the number of touchdowns. Touchdowns are worth 7 points after a one-point conversion, and field goals are worth 3 points.
A.{t=2f+17t+3f=68C.{t=2f+13t+7f=68B.{f=2t+17t+3f=68D.{f=2t+13t+7f=68
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{"type":"text","form":{"type":"equation","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"","description":[{"id":0,"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.9580078125em;vertical-align:-0.2080078125em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Regular\">Field<\/span><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">goals<\/span><\/span><\/span><\/span><\/span>"},{"id":1,"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.759765625em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Regular\">Touchdowns<\/span><\/span><\/span><\/span><\/span>"}]},"formTextBefore":null,"formTextAfter":null,"answer":{"text":[{"id":0,"text":"11"},{"id":1,"text":"5"}],"noSolution":false,"infiniteSolution":false}}
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We know that the Panthers scored 68 points. We also know that each touchdown t is worth 7 points, after a one-point conversion, which can be expressed as 7t. Furthermore, each field goal f is worth 3 points, which we can write as 3f. With this information, we can write our first equation. 7t+4f=68 We are also told that the number of field goals was 1 more than twice the number of touchdowns. Let's write an equation to represent this statement. f=2t+1 The two equations we wrote form the system of linear equations that corresponds to choice B. cc A. t=2f+1 7t+3f=68 & B. f=2t+1 7t+3f=68 [1.75em] C. t=2f+1 3t+7f=68 &D. f=2t+1 3t+7f=68
To find the number of touchdowns and field goals that the Panthers scored last night, we will solve the system we wrote in Part A. Since f is already isolated in the first equation, we can use the Substitution Method.
The number of touchdowns the Panthers scored is 5. Now we can substitute this number for t into the first equation and simplify to determine the value of f.
The Panthers scored 11 field goals.
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LaShay is climbing down a ladder from her roof, while Tiffaniqua is climbing up a second identical ladder that is placed next to the first ladder. Each ladder has 30 rungs. Their friend Davontay recorded the following information.
- LaShay went down 2 rungs every second.
- Tiffaniqua went up 1 rung every second.
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Let R be the rung number from the base of the ladder and t be the time in seconds. We know that LaShay climbs down the 30-rung ladder at a speed of 2 rungs per second. At the same time, Tiffaniqua climbs up an identical ladder at a speed of 1 rung per second.
We will write an equation for each person, then form a system of equations. LaShay starts at the 30^(th) rung and steps down 2 rungs each second. With this information, we can write our first equation. R = 30 - 2t Tiffaniqua starts from the base of the ladder and climbs up 1 rung every second. Let's write a second equation. R= 1t+0 ⇔ R=t If we combine these equations, we can form a linear system that we can solve by using the Substitution Method.
Tiffaniqua and LaShay will meet on the 10^(th) rung from the base.
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The school library sells hardcover books for $5 each and paperback books for $3 each. If Ramsha spends $38 for 10 books, how many paperback books did she buy?
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To write a system of equations that represents the situation, we need to first assign variables to each item. We can say that the number of hardcover books is h and the number of paperback books is p. We know that Ramsha bought 10 books in an unknown combination of hardcovers and paperbacks. This can be illustrated by the following equation. h+p=10 We also know that each hardcover book costs $ 5 and that each paperback book costs $ 3. Furthermore, Ramsha spent a total of $38. This can be shown by a second equation. 5h+3p=38 The two equations above form a system of equations. h+p=10 & (I) 5h+3p=38 & (II) We can most easily use the Substitution Method to solve this system, since we can easily isolate either variable in the first equation. Let's begin by isolating h.
Ramsha bought 6 paperback books. We can find the value of h by substituting p=6 in the first equation. However, since we are only asked how many paperback books Ramsha bought, we do not need the value of h.
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Modeling with Systems of Linear Equations
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