equations and problem solving
DESCRIPTION
Equations and Problem Solving. Using Algebra to solve Word Problems: Counting Problems and Mixture Problems. Definition. The sum of 3 consecutive integers is 147. Find the integers. + n+2. + n+1. n. n + n+1 + n+2 =147 n + n + 1 + n + 2=147 3n + 3=147 3n=144 n=48. - PowerPoint PPT PresentationTRANSCRIPT
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