Exercise 20 Page 227

Let's start by defining the variables. Let p be the number of personal tubes and c the number of cooler tubes. Since we are told that the group rents a total of 15 tubes, the sum of p and c must be 15. p+c=15 We also know that the group spends $270 in renting the tubes. Additionally, the cost of a one-person tube is $20, and the cost of renting a cooler tube is $12.50. We can write our second equation using this information. 20p+12.5c=270 The equations combined together for a system of linear equations. p+c=15 & (I) 20p+12.5c=270 & (II) To find how many of each type of tube the group rents, we have to solve for p and c. To do so, we will use the Substitution Method. Let's start by isolating the p-variable in Equation (I).

p+c=15 20p+12.5c=270

SubEqn

(I): LHS-c=RHS-c

p=15-c 20p+12.5c=270

Now we can substitute 15-c for p in Equation (II).

p=15-c 20p+12.5c=270

Substitute

(II):p= 15-c

p=15-c 20( 15-c)+12.5c=270

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Solve for c

Distr

(II): Distribute 20

p=15-c 300-20c+12.5c=270

AddTerms

(II): Add terms

p=15-c 300-7.5c=270

SubEqn

(II): LHS-300=RHS-300

p=15-c - 7.5c=- 30

DivEqn

(II): .LHS /(- 7.5).=.RHS /(- 7.5).

p=15-c c=4

Now we can find p by substituting 12 for c in Equation(I).

p=15-c c=4

Substitute

(I): c= 4

p=15- 4 c=4

SubTerm

(I): Subtract term

p=11 c=4

The group rents 11 personal tubes and 4 cooler tubes.