Saturday, October 31, 2015

Solving simultaneous equations


Here is a link to practice problems for simultaneous equations. The coin problems all use elimination instead of substitution. Here are the same coin problems using substitution, with the answers in the comment below.

a) 100 coins, all quarters and pennies, total = $13.72

b) 100 tickets, all children ($6) and adult ($12), total =$912

c) 100 coins, all quarters and nickels, total = $8.20

d) 100 coins, all dimes and pennies, total = $3.79

e) 100 coins, all dimes and nickels, total = $7.85

f) 100 coins, all quarters and dimes, total = $14.65
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1 comment:

  1. Prof. HubbardOctober 31, 2015 at 9:00 AM

    a) 100 coins, all quarters and pennies, total = $13.72

    q + p = 100
    25q + p = 1372

    Use the first equation to get p = 100 - q
    The second equation becomes

    25q + 100 - q = 1372
    24q + 100 = 1372
    24q = 1272
    q = 1272/24 = 53
    p = 100 - 53 = 47

    b) 100 tickets, all children ($6) and adult ($12), total =$912

    c + a = 100
    6c + 12a = 912

    Use the first equation to get c = 100 - a
    The second equation becomes

    6(100 - a) + 12a = 912
    600 - 6a + 12a = 912
    600 + 6a = 912
    6a = 312
    a = 312/6 = 52
    c = 100 - 52 = 48


    c) 100 coins, all quarters and nickels, total = $8.20

    q + n = 100
    25q + 5n = 820

    Use the first equation to get n = 100 - q
    The second equation becomes

    25q + 5(100 - q) = 820
    25q + 500 - 5q = 820
    20q + 500 = 820
    20q = 320
    q = 320/20 = 16
    c = 100 - 16 = 84

    d) 100 coins, all dimes and pennies, total = $3.79

    d + p = 100
    10d + p = 379

    Use the first equation to get p = 100 - d
    The second equation becomes

    10d + 100 - d = 379
    9d + 100 = 379
    9d = 279
    d = 279/9 = 31
    p = 100 - 31 = 69


    e) 100 coins, all dimes and nickels, total = $7.85
    d + n = 100
    10d + 5n = 785

    Use the first equation to get n = 100 - d
    The second equation becomes

    10d + 5(100 - d) = 785
    10d + 500 - 5d = 785
    5d + 500 = 785
    5d = 285
    d = 285/5 = 55
    n = 100 - 55 = 45


    f) 100 coins, all quarters and dimes, total = $14.65

    q + d = 100
    25q + 10d = 1465

    Use the first equation to get d = 100 - q
    The second equation becomes

    25q + 10(100 - q) = 1465
    25q + 1000 - 10q = 1465
    15q + 1000 = 1465
    15q = 465
    q = 465/15 = 31
    d = 100 - 31 = 69

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