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Core Connections Integrated I, 2013 View details

2. Section 8.2

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Exercise 114 Page 480

Practice makes perfect

a From the exercise we know that the Lee's have three children. We will call the age of these children, from youngest to oldest,

x, y

and

z .

Using the information in the exercise, we can write three equations.

T h e o l d e s t is twice as old as the youngest z = 2 x T h e m i d d l e c h i l d is 5 years older than the youngest y = x + 5 the sum of the ages is 57 x + y + z = 57

If we combine these functions we get a system of equations, which we can solve by using the Substitution Method.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 2 x y = x + 5 x + y + z = 57 (I) (II) (III)

SubstituteII

$(III):$ $z = 2 x$ , $y = x + 5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 2 x y = x + 5 x + (x + 5) + 2 x = 57

▼

(III):

Solve for

x

RemovePar

$(III):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 2 x y = x + 5 x + x + 5 + 2 x = 57

AddTerms

$(III):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 2 x y = x + 5 4 x + 5 = 57

SubEqn

$(III):$ $LHS - 5 = RHS - 5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 2 x y = x + 5 4 x = 52

DivEqn

$(III):$ $LHS / 4 = RHS / 4$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 2 x y = x + 5 x = 13

When we know the value of

x

we can substitute this into the first and second equation to find the age of the remaining kids.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 2 x y = x + 5 x = 13

$(I), (II):$ $x = 13$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 2 (13) y = 13 + 5 x = 13

Multiply

$(I):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 26 y = 13 + 5 x = 13

AddTerms

$(II):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 26 y = 18 x = 13

The ages of the kids are

26,

18

and

13 .

b Before we can solve the exercise we will introduce a few variables.

x y = number of days = number of insects

In the garden there are currently

105

ladybugs. Every month, the ladybugs increase at

2

ladybugs per month. With this information we can write an equation that describes the amount of ladybugs in the garden.

y = 105 + 2 x

Also, there are

175

aphids in the garden, and this amount is decreasing at

3

aphids every month. With this information, we can write a second equation that describes the amount of aphids in the garden.

y = 175 - 3 x

If we combine these equations we get a system of equations, which we can solve with the Substitution Method.

{y = 105 + 2 x y = 175 - 3 x (I) (II)

▼

Solve by substitution

Substitute

$(I):$ $y = 175 - 3 x$

{175 - 3 x = 105 + 2 x y = 175 - 3 x

SubEqn

$(I):$ $LHS - 175 = RHS - 175$

{- 3 x = - 70 + 2 x y = 175 - 3 x

SubEqn

$(I):$ $LHS - 2 x = RHS - 2 x$

{- 5 x = - 70 y = 175 - 3 x

MultEqn

$(I):$ $LHS \cdot (- 1) = RHS \cdot (- 1)$

{5 x = 70 y = 175 - 3 x

DivEqn

$(I):$ $LHS / 5 = RHS / 5$

{x = 14 y = 175 - 3 x

After

14

days the number of ladybugs and aphids is the same. Note that we do not have to bother solving for

y,

as we were only asked when the number of insects is the same. We do not have to figure out what that number is.

c We know the price for

5

pounds of tomatoes. To find the price of

3

pounds we first need to determine the unit price, which is the price for

1

pound. If

5

pounds costs

$ 8,

then the cost of

1

pound is the quotient of

8

and

5 .

\frac{8}{5} = 1.6

The unit price is

$ 1.60 .

Therefore, the price of

3

pounds must be

3 (1.60) = $ 4.80 .

d Let's label the number of adult tickets

a

and the number of student tickets

s .

30

more student tickets were sold than adult tickets, we can write the following equation.

s = 30 + a

Also, the cost of an adult ticket is

$ 5

and the cost of a student ticket is

$ 3 .

The total sales of both types of tickets amount to

$ 1770 .

With this information, we can write the following equation.

5 a + 3 s = 1770

If we combine these equations we get a system of equations, which we will solve with the Substitution Method.

{s = 30 + a 5 a + 3 s = 1770 (I) (II)

▼

Solve by substitution

Substitute

$(II):$ $s = 30 + a$

{s = 30 + a 5 a + 3 (30 + a) = 1770

Distr

$(II):$ Distribute $3$

{s = 30 + a 5 a + 90 + 3 a = 1770

AddTerms

$(II):$ Add terms

{s = 30 + a 8 a + 90 = 1770

SubEqn

$(II):$ $LHS - 90 = RHS - 90$

{s = 30 + a 8 a = 1680

DivEqn

$(II):$ $LHS / 8 = RHS / 8$

{s = 30 + a a = 210

Substitute

$(I):$ $a = 210$

{s = 240 a = 210

The school sold

240

student tickets and

210

adult tickets.

Subchapter links

8.2.1 Curve Fitting p.471-472

100

101

102

8.2.2 Curved Best-Fit Models p.478-480

108

109

110

111

112

113

114

8.2.3 Solving a System of Exponential Functions Graphically p.483-484

119

120

121

122

123

124

125

Exercise 114 Page 480

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