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4.2B Word Problems - Solving Linear System by
Substitution
Previously when solving word problems you were restricted to one variable. Today you will solve word problems using more than one variable. If you use more than one variable, you will need to write a system of linear equations.
1) Write two sets of labels, if necessary (one set for number, one set for value, weight etc.)
2) Write two verbal models. (Translate from sentences)
3) Write two algebraic models (equations).
4) Solve the system of equations.
5) Write a sentence and check your solution in the word problem.
Solving Word Problems Using A Linear System
Jenny has 8 moving boxes that she can use to pack for college. Each box can hold 15 pounds of clothing or 60 pounds of books. If Jenny is moving 255 pounds how many boxes of each type are there?
Let c = # of clothing boxes
Jenny has 8 moving boxes that she can use to pack for college. Each box can hold 15 pounds of clothing or 60 pounds of books. If Jenny is moving 255 pounds how many boxes of each type are there?
Let 15c =
Let b = # of book boxes
Number Labels. Weight Labels Let 60b
=
weight of the clothing boxes
weight of book boxes
c + b = 8
Verbal Model (to represent the number of boxes)
# of clothing boxes + # of book boxes = Total #Equati
on
15c + 60b = 255
Verbal Model (to represent the weight of boxes) Weight of
clothing boxesEquatio
n
Weight of book boxes
Total Weight+ =
Solve the linear system.
255 60c15 Solve the linear system.
Choose one equation and isolate one of the variables.
8 cb
Substitute the expression into the other equation and solve.Substitute the solved value into one of the original equations and solve.
5c225c45
255480c45255480c60c15
Sentence.
8bc
3b8b5
Jenny has 3 book boxes and 5 clothing boxes .
Could the answer be 2 book boxes and 6 clothing boxes? Check the solution in the word problem.
8 c 255b 60c 15 and 8bc
Word problems require word answers.
Let a = # of adults
Example 1 In one day the museum collected $1590 from 321 people. The price of admission is $6 for an adult and $4 for a child. How many adults and how many children were admitted to the museum?
Let 6a =
Let c = # of children
Number Labels. Value Labels Let 4c
=
value of adult tickets
value of children’s tickets
Let a = # of adults
Example 1 In one day the museum collected $1590 from 321 people. The price of admission is $6 for an adult and $4 for a child. How many adults and how many children were admitted to the museum?
Let 6a =
Let c = # of children
Number Labels. Value Labels Let 4c
=
value of adult tickets
value of children’s tickets
a + c = 321
Verbal Model (to represent the number of people)
# of adultsEquation
# of children
Total # of people
+ =
6a + 4c = 1590
Verbal Model (to represent the value of tickets/admission) Value of
adult ticketsEquation
Value of children’s tickets
Total Value
+ =
Solve the linear system.
Solve the linear system.
321 ca
168c336c2
15901926c21590c41926c6
321ca
153a321168a
The museum admitted 153 adults and 168 children.
1590c4 6 321 c 1590c 4a 6 and 321ca
Are there other values that will total 321? How do you know if you have the correct combination?
Let a = # of Stock A
Example 2 An investor bought 225 shares of stock. Stock A was purchased at $50 per share and Stock B at $75 per share. If $13,750 worth of stock was purchased, how many shares of each kind did the investor buy?
Let 50a = value of Stock A
Let b = # of Stock B
Number LabelsValue Labels Let 75b = value of
Stock B
Let a = # of Stock A
Example 2 An investor bought 225 shares of stock. Stock A was purchased at $50 per share and Stock B at $75 per share. If $13,750 worth of stock was purchased, how many shares of each kind did the investor buy?
Let 50a = value of Stock A
Let b = # of Stock B
Number LabelsValue Labels Let 75b = value of
Stock B
a + b = 225
Verbal Model (to represent the number of stock) # of Stock
AEquation
# of Stock B
Total # of stock
+ =
50a + 75b = 13,750
Verbal Model (to represent the value of stock)
Value of Stock A + Value of Stock B = Total Value of stockEquation
Solve the linear system.
Solve the linear system.
225 ba
100b2500b25
750,13250,11b25750,13b75250,11b50
225ba
125a225100a
The investor purchased 125 shares of Stock A and 100 shares of Stock B.
750,13b75 05
225ba
225 b
750,13b75a50 and
110w22 l
110w2) (2
Example 3 The length of a rectangle is 1 m more than twice its width. If the perimeter is 110 m, find the dimensions.
Length
let w = widthlet l = length
Formula
length
length
widthwidth
2 lengths + 2 widths = perimeter 1w2 l
18w108w61102w6110w22w4
The width is 18 m and the length is 37 m.
37136
1182
=
2 widths + 1=
1w2
Practice Problems
1. A sightseeing boat charges $5 for children and $8 for adults. On its first trip of the day, it collected $439 for 71 paying passengers. How many children and how many adults were there?
2. The length of a rectangle is 12 inches more than twice its width. If the perimeter is 90 inches, what are the dimensions of the rectangle?
Practice Problems
1. A sightseeing boat charges $5 for children and $8 for adults. On its first trip of the day, it collected $439 for 71 paying passengers. How many children and how many adults were there? There were 43 children and 28 adults on the sightseeing trip.
2. The length of a rectangle is 12 inches more than twice its width. If the perimeter is 90 inches, what are the dimensions of the rectangle? The width is 11 inches and the length is 34 inches.
7-A3 Pages 399-401 # 24–33, 48–52.
