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Algebra 1 PowerPoint Presentation

Algebra 1

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Algebra 1

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  1. Algebra 1 Coin Word Problems

  2. Stan’s pocket is filled with nickels and dimes. He has 6 more dimes than nickels. All together Stan has $1.20 in coins. How many nickels and dimes does he have? -Let’s start with the sentence “He has 6 more dimes than nickels.” -Ask yourself “Which coin does he have more of?” -The answer is dimes. d = n + 6 Quantity Equation

  3. All together Stan has $1.20 in coins. • Each coin has a different value. • Each nickel is worth 5 cents. Thus 5n is the amount all his nickels are worth. • Each dime is worth 10 cents. Thus 10d is the amount all his dimes are worth. • $1.20 is worth 120 cents… • 5n + 10d = 120 The value equation.

  4. Substitute and solve • 10d + 5n = 120 and d = (n + 6) • 10(n + 6) + 5n = 120 • 10n + 60 + 5n = 120 • 15n + 60 = 120 • 15n = 60 • n = 4

  5. Complete the Problem • He has 4 nickels. • d = n + 6 • d = 4 + 6 • d = 10 • Stan has 4 nickels and 6 dimes.

  6. Check • 10 dimes is worth 10(.10) = $1.00 • 4 nickels is worth 4(.05) = $0.20 Total = $1.20

  7. 4) She gave the sitter 6 more nickels than dimes. • She has more… • Nickels. • n = d + 6 Quantity Equation. • She has 3 times as many quarters as dimes. • She has more… • Quarters. • q =3d Quantity Equation.

  8. All together Mrs. Ryan paid the babysitter $7.50. • Total nickel value in cents… • 5n • Total dime value in cents… • 10d • Total quarter value in cents… • 25q • 5n + 10d + 25q = 750

  9. Substitute and Solve • n = d + 6 q = 3d • 5n + 10d + 25q = 750 • 5(d + 6) + 10d + 25(3d) = 750 • 5d + 30 + 10d + 75d = 750 • 90d + 30 = 750 • 90d = 720 • d = 8 • She paid 8 dimes.

  10. Solve for Nickels and Quarters • d = 8 • n = d + 6 • n = 8 + 6 • n = 14 • q = 3d • q = 3(8) • q =24

  11. Check • 8 dimes is worth 8($0.10) = $0.80 • 14 nickels is worth 14($0.05) = $0.70 • 24 quarters is worth 24($0.25) = $6.00 Total = $7.50