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3 September, 00:48

Which equation can be used to find two numbers such that twice the first equals three times the second and three times their difference exceeds twice their difference by 13?

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  1. 3 September, 01:14
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    Let the numbers be a and b and b>a

    2a=3b and 3 (b-a) = 2 (b-a) + 13

    Solving the first for a:

    2a=3b

    a=3b/2, now using this value of a in the second equation gives you:

    3 (b-3b/2) = 2 (b-3b/2) + 13 upon performing indicated operations.

    3b-9b/2=2b-6b/2+13 making all terms have a common denominator of 2

    (6b-9b) / 2 = (4b-6b+26) / 2 multiplying the whole equation by 2

    6b-9b=4b-6b+26 combining like terms

    -3b=-2b+26 adding 2b to both sides

    -b=26 dividing both sides by - 1

    b=-26, since a=3b/2

    a=3 (-26) / 2

    a=-39

    So the numbers are - 39 and - 26

    check ...

    2a=3b becomes:

    2 (-39) = 3 (-26)

    -78=-78

    3 (b-a) = 2 (b-a) + 13 becomes:

    3 (-26--39) = 2 (-26--39) + 13

    3 (13) = 2 (13) + 13

    39=26+13

    39=39

    correct : P
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