Equations and Problem Equations and Problem SolvingSolving

Using Algebra to solve Word Problems: Counting Problems and Mixture Problems

DefinitionDefinition

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CONSECUTIVE INTEGERSCONSECUTIVE INTEGERSIntegers that differ by one. The integers 50 and 51

are consecutive and so are -10 and -9

Let n = the first integerThen n+1 = the second integerAnd n+2 = the third integer

Let n = the first integerThen n+1 = the second integerAnd n+2 = the third integer

n + n+1 + n+2 = 147n + n + 1 + n + 2 = 147

3n + 3 = 1473n = 144

n = 48

The sum of 3 consecutive integers is 147. Find the integers.

The sum of 3 consecutive integers is 147. Find the integers.

Let n = 48Then n+1 = 49And n+2 = 50

Let n = 48Then n+1 = 49And n+2 = 50

+ n+2n

+ n+1

ExamplesExamples• The sum of 3 consecutive integers is 72. Find

the integers.

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Let n = the first integerThen n+1 = the second integerAnd n+2 = the third integer

n + n+1 + n+2 = 723n + 3 = 72

3n = 69n = 23

Let n = 23Then n+1 = 24And n+2 = 25

ExamplesExamples• The sum of 3 consecutive integers is 915. Find

the integers.

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Let n = the first integerThen n+1 = the second integerAnd n+2 = the third integer

n + n+1 + n+2 = 9153n + 3 = 915

3n = 912n = 304

Let n = 304Then n+1 = 305And n+2 = 306

Mixture ProblemsMixture Problems• John has $1.70, all in dimes and nickels. He

has a total of 22 coins. How many of each kind does he have?

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Step 1:Step 1:Assigning variablesStep 1:Step 1:Assigning variables d = dimesd = dimes n = nickelsn = nickels

Step 2:Step 2:Write algebraic equationStep 2:Step 2:Write algebraic equation d + n = 22d + n = 22

Step 3:Step 3:Write value equationStep 3:Step 3:Write value equation 0.10d + 0.05n = 1.700.10d + 0.05n = 1.70

Mixture ProblemsMixture Problems• John has $1.70, all in dimes and nickels. He

has a total of 22 coins. How many of each kind does he have?

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Step 4:Step 4:Solve the mixture problemStep 4:Step 4:Solve the mixture problem

0.10d + 0.05n = 1.7010d + 5n = 170

0.10d + 0.05n = 1.7010d + 5n = 170

d + n = 2210d + 10n = 220

d + n = 2210d + 10n = 220

Mixture ProblemsMixture Problems• John has $1.70, all in dimes and nickels. He

has a total of 22 coins. How many of each kind does he have?

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Step 4:Step 4:Solve the mixture problemStep 4:Step 4:Solve the mixture problem

10d + 5n = 17010d + 10n = -220

-5n = -50

n = 5

10d + 5n = 17010d + 10n = -220

-5n = -50

n = 5

Change the sign of the 2nd equation

Mixture ProblemsMixture Problems• John has $1.70, all in dimes and nickels. He

has a total of 22 coins. How many of each kind does he have?

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Step 4:Step 4:Solve the mixture problemStep 4:Step 4:Solve the mixture problem

10d + 5n = 17010d + 5(10) = 170

10d + 50 = 170

10d = 120d = 12

10d + 5n = 17010d + 5(10) = 170

10d + 50 = 170

10d = 120d = 12

Substitute “n” with 10

Warm UpWarm Up

The sum of 3 consecutive integers is 60. What are the values of the 3 integers?

The 3 consecutive numbers are 29, 30, and 31.

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Mixture ProblemMixture ProblemTickets to a movie cost $5.00 for adults and $3.00 for children. If tickets were bought for 50 people for a total of $196 how many adult tickets were sold and how many children tickets were sold?

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1: Assigning variables1: Assigning variables a = adultsa = adults c = childrenc = children

2: Write algebraic equation2: Write algebraic equation a + c = 50a + c = 50

3: Write value equation3: Write value equation 5a + 3c = 1965a + 3c = 196

Mixture ProblemMixture Problem

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4: Solve the mixture problem4: Solve the mixture problem

a + c = 50-5a – 5c = -2505a + 3c = 196

-2c = -54

c = 27

a + c = 50-5a – 5c = -2505a + 3c = 196

-2c = -54

c = 27-5a – 5c = -250

-5a – 5(27) = -250-5a – 135 = -250

-5a = -115a = 23

-5a – 5c = -250-5a – 5(27) = -250

-5a – 135 = -250

-5a = -115a = 23

Mixture ProblemMixture Problem

Tickets to a movie cost $4.00 for adults and $2.00 for children. If tickets were bought for 80 people for a total

of $230 how many adult tickets were sold and how many children tickets were sold?

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c = 45a = 35

HomeworkHomework

Amy has 32 coins consisting of dimes and quarters. If Amy has a total of $5 in her pocket,

how many of each coin are there?

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quarters = 12dimes = 20