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8 October, 14:22

how to solve 0.01x+0.07y=0.22 and 0.03x-0.05y=0.14 by substitution

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  1. 8 October, 17:47
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    0.01x + 0.07y = 0.22

    0.03x - 0.05y = 0.14

    x + 7y = 22

    3x - 5y = 14

    x = 22 - 7y

    3 * (22-7y) - 5y = 14

    66 - 21y - 5y = 14

    -26y = - 52

    y = 2

    x + 14 = 22

    x = 8
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