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Warm-Up

5 minutes. Warm-Up. Beth and Chris drove a total of 233 miles in 5.6 hours. Beth drove the first part of the trip and averaged 45 miles per hour. Chris drove the second part of the trip and averaged 35 miles per hour. For what length of the time did Beth drive?. Digit and Coin Problems.

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Warm-Up

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  1. 5 minutes Warm-Up Beth and Chris drove a total of 233 miles in 5.6 hours. Beth drove the first part of the trip and averaged 45 miles per hour. Chris drove the second part of the trip and averaged 35 miles per hour. For what length of the time did Beth drive?

  2. Digit and Coin Problems Objectives: To use systems of equations to solve digit and coin problems

  3. Example 1 The sum of the digits of a two-digit number is 10. If the digits are reversed, the new number is 36 less than the original number. Find the original number. Let x = the tens digit 10x = the value of tens digit Let y = the ones digit 1y = the value of ones digit 2x - 4= 10 x + y = 10 Tens digit + ones digit = 10 2x= 14 When reversed value of tens digit=10y and ones digit = x 10y + x = 10x + y - 36 x= 7 Simplify 9y = 9x - 36 y = x - 4 y = x - 4 y = 7 - 4 1st equation x + y = 10 y = 3 x +(x – 4)= 10 substitute 2x - 4= 10 73

  4. Practice The sum of the digits of a two-digit number is 5. If the digits are reversed, the new number is 27 more than the original number. Find the original number.

  5. Example 2 A collection of nickels and dimes is worth $3.95. There are 8 more dimes than nickels. How many dimes and how many nickels are there? Let n be the number of nickels. Let .05n =value of nickels. Let d be the number of dimes. Let .10d =value of dimes. 0.05n + 0.10d = 3.95 Value of nickels + value of dimes = $3.95 d = 8 + n The number of dimes is 8 more than the number of nickels Substitute 0.05n + 0.10(8 + n) = 3.95 Substitute d = 8 + 21 Simplify 0.05n + 0.80 + 0.10n = 3.95 d = 29 5n + 80 + 10n = 395 21 nickels 80 + 15n = 395 29 dimes 15n = 315 n = 21

  6. Practice Rob has $2.85 in nickels and dimes. He has twelve more nickels than dimes. How many of each coin does he have?

  7. Example 3 There were 166 paid admissions to a game. The price was $2 for adults and $0.75 for children. The amount taken in was $293.25. How many adults and how many children attended? Let a be the number of adults who attended Let c be the number of children who attended a + c = 166 Number of adults + children = 166 2a + 0.75c = 293.25 $2 times the # of adults + $0.75 times the # of children = $293.25 a + c = 166 Isolate one of the variables in the first equation a = 166 - c a + c = 166 1st equation 2(166 – c)+ 0.75c = 293.25 Substitute a + 31 = 166 Substitute 332 – 2c + 0.75c = 293.25 a = 135 332 - 1.25c = 293.25 135 adults - 1.25c = -38.75 c = 31 31 children

  8. Practice The attendance at a school concert was 578. Admission cost $2 for adults and $1.50 for children. The receipts totaled $985.00. How many adults and how many children attended the concert?

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