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
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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
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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.
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=47 & (I) 5x+4y=200 & (II)
Finally, the above system will be solved. For simplicity, the Substitution Method will be used.
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 5x+4y=200
SubEqn
(I): LHS-x=RHS-x
y=47-x 5x+4y=200
Substitute
(II): y= 47-x
y=47-x 5x+4( 47-x)=200
y=47-x x=12
Extra
The answer can also be found by solving a system considering the extra dollar.
x+y=47 & (I) 5x+4y=201 & (II)
The Substitution Method will be used.
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=47 5x+4y=201
▼
Solve by substitution
SubEqn
(I): LHS-x=RHS-x
y=47-x 5x+4y=201
Substitute
(II): y= 47-x
y=47-x 5x+4( 47-x)=201
Distr
(II): Distribute 4
y=47-x 5x+188-4x=201
SubTerm
(II): Subtract term
y=47-x x+188=201
SubEqn
(II): LHS-188=RHS-188
y=47-x x=13
Substitute
(I): x= 13
y=47- 13 x=13
SubTerm
(I): Subtract term
y=34 x=13
Example
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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.
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=50 & (I) 3.11x+2.5y=138.42 & (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.
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 3.11x+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=125 3.11x+2.5y=138.42
2.5x+2.5y=125 3.11x+2.5y=138.42
SysEqnSub
(II): Subtract (I)
2.5x+2.5y=125 3.11x+2.5y-( 2.5x+2.5y)=138.42- 125
2.5x+2.5y=125 x=22
Example
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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.
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=13 & (I) 12x+8y=128 & (II)
For simplicity, the system will be solved by graphing. To do so, both linear equations must first be written in slope-intercept form.
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 12x+8y=128
▼
(I), (II): Write in slope-intercept form
SubEqn
(I): LHS-x=RHS-x
y=- x+13 12x+8y=128
SubEqn
(II): LHS-12x=RHS-12x
y=- x+13 8y=- 12x+128
DivEqn
(II): .LHS /8.=.RHS /8.
y=- x+13 y= - 12x+1288
WriteSumFrac
(II): Write as a sum of fractions
y=- x+13 y= - 12x8+ 1288
MovePartNumRight
(II): a* b/c=a/c* b
y=- x+13 y= - 128x+ 1288
MoveNegNumToFrac
(II): Put minus sign in front of fraction
y=- x+13 y=- 128x+ 1288
ReduceFrac
(II): a/b=.a /4./.b /4.
y=- x+13 y=- 32x+ 1288
CalcQuot
(II): Calculate quotient
y=- x+13 y=- 32x+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
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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.
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: & 150x+10x=160x Child: & 75y+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≤ 8 & (I) 160x+85y+200≤1200 & (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.
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 160x+85y+200 = 1200
▼
(I), (II): Write in slope-intercept form
SubEqn
(I): LHS-x=RHS-x
y=- x+8 160x+85y+200=1200
SubEqn
(II): LHS-200=RHS-200
y=- x+8 160x+85y=1000
SubEqn
(II): LHS-160x=RHS-160x
y=- x+8 85y=- 160x+1000
DivEqn
(II): .LHS /85.=.RHS /85.
y=- x+8 y= - 160x+100085
WriteSumFrac
(II): Write as a sum of fractions
y=- x+8 y= - 160x85+ 100085
MoveNegNumToFrac
(II): Put minus sign in front of fraction
y=- x+8 y=- 160x85+ 100085
MovePartNumRight
(II): a* b/c=a/c* b
y=- x+8 y=- 16085x+ 100085
ReduceFrac
(II): a/b=.a /5./.b /5.
y=- x+8 y=- 3217x+ 20017
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
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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.
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=18 & (I) 4x+2y=40 & (II)
Finally, the system will be solved. For simplicity, the Substitution Method will be used.
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 4x+2y=50
▼
(I), (II): Solve by substitution
SubEqn
(I): LHS-x=RHS-x
y=- x+18 4x+2y=50
Substitute
(II): y= - x+18
y=- x+18 4x+2( - x+18)=50
Distr
(II): Distribute 2
y=- x+18 4x-2x+36=50
SubTerm
(II): Subtract term
y=- x+18 2x+36=50
SubEqn
(II): LHS-36=RHS-36
y=- x+18 2x=14
DivEqn
(II): .LHS /2.=.RHS /2.
y=- x+18 x=7
Substitute
(I): x= 7
y=- 7+18 x=7
AddTerms
(I): Add terms
y=11 x=7
Modeling with Systems of Linear Equations
Exercise 3.1
A company makes framed mirrors of two different sizes. The cost of making a small framed mirror is $59 and the cost of making a large framed mirror is $81.
Each framed mirror contains the mirror itself and a wooden frame that has two horizontal parts and two vertical parts. The edges of these frames have been sawed at a 45^(∘)-angle and overlap the mirror by 2 centimeters.
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To write a general expression for the cost, we need to figure out two things.
- The price per square meter of the mirror, which we will label x.
- The price per meter of the wooden frame, which we will label y.
We know the cost of the smaller and bigger framed mirrors. Using this information, we can write a system of equations that describes the price per square meter of the mirror and the price per meter of the wooden frame.
Smaller Framed Mirror
The frame is 5 centimeters wide and overlaps the mirror by 2 centimeters. Therefore, to determine the length and width of the mirror, we subtract 3 centimeters from each side of the frame.
The length and width of the smaller mirror are 40 and 30 centimeters, respectively. Recall that the price is given per square meter. Therefore, we must convert centimeters into meters. We can do this by multiplying each measure by the conversion factor 1m100cm.
| Centimeters | Convert | Meters | |
|---|---|---|---|
| Length | 40 | 40cm* 1m/100cm | 0.4 |
| Width | 30 | 30cm* 1m/100cm | 0.3 |
Next, we can find the area, in square meters, of the smaller mirror. To do so, we multiply the length by the width. Also, recalling that the cost of one square meter is x, we can write an expression for the cost of the mirror. cc Area & Cost 0.4(0.3)= 0.12 m^2 & $0.12x Let's now find an expression for the cost of the frame. To do this, we need to consider four pieces of wood, two for the width and two for the length.
Again, we need to rewrite these measures into meters.
| Centimeters | Convert | Meters |
|---|---|---|
| 46 | 46cm* 1m/100cm | 0.46 |
| 36 | 36cm* 1m/100cm | 0.36 |
Next, we will add the lengths of the pieces to obtain the total meters of wooden frame we need. Also, considering that the cost of one meter is y, we can write an expression for the cost of the frame. cc Frame & Cost 2(0.46)+2(0.36)=1.64 m & $1.64y Finally, by adding the costs of the mirror and the wooden frame, we can write an expression for the total cost of the smaller framed mirror, which is $ 59. 0.12x+1.64y=59
Larger Framed Mirror
We will repeat the same process for the bigger mirror. Just like before, we need to subtract 6 centimeters from the length and width of the frame to obtain dimensions of the mirror. Length:& 56-6=50cm Width:& 46-6=40cm Next, we convert these measures to meters by multiplying by the conversion factor 1m100cm.
| Centimeters | Convert | Meters | |
|---|---|---|---|
| Length | 50 | 50cm* 1m/100cm | 0.5 |
| Width | 40 | 40cm* 1m/100cm | 0.4 |
Now we can find the area, in square meters, of the bigger mirror. To do so, we multiply the length by the width. Also, recalling that the cost of one square meter is x, we can write an expression for the cost of the mirror. cc Area & Cost 0.5(0.4)= 0.2 m^2 & $0.2x Just like we did with the smaller framed mirror, to determine the total length of the frame we convert the measures to meters and add the length of the four pieces.
| Centimeters | Convert | Meters |
|---|---|---|
| 56 | 56cm* 1m/100cm | 0.56 |
| 46 | 46cm* 1m/100cm | 0.46 |
| Sum | 2(0.56)+2(0.46)=2.04 | |
Therefore, the wooden frame costs $2.04y. If we add the cost of the mirror and the cost of the frame we obtain the total cost of the bigger framed mirror, which is $81. 0.2x+2.04 y=81
System of Equations
We can combine these two equations to get a system of equations. If we solve it, we will find the price per square meter of the mirror itself and the price per meter for the wooden frame. Let's use the Elimination Method.
The mirror costs $150 per square meter and the wooden frame costs $25 per meter.
General Expression
Finally, we can write a general expression for a framed mirror that is a meters wide and b meters long. As already mentioned, the frame overlaps 2 centimeters of each side of the mirror. Therefore, to get the area of the mirror, we subtract 6 centimeters, which is 6* 1100=0.06 meters, from the length and width of the frame.
We can find the total price for the mirror by multiplying the price per square meter $ 150 by the area of the mirror. Price of the Mirror 150(a-0.06)(b-0.06) We determine the cost of the wooden frame by multiplying the price per meter $ 25 by the total length of the frame. Price of the Wooden Frame 25(2a+2b) Finally, we can write an expression for the total cost of a framed mirror that is a meters long by b meters wide by adding these two costs.
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Modeling with Systems of Linear Equations
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