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19 December, 10:59

The annual interest on a $14,000 investment exceeds the interest earned on a $7000 investment by $595. The $14,000 is invested at a 0.5% higher rate of interest than the $7000. What is the interest rate of each investment?

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  1. 19 December, 11:08
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    x = 8.00 Interest rate on $14000

    y = 7.50 Interest rate on $7000

    Step-by-step explanation:

    Let interest rate of $14000 be x%

    and Interest rate for $7000 be y %

    According to the first condition

    14000 * x% - 7000 * y% = 595

    multiply by 100

    14000x-7000y = 59500

    /700

    20x-10y=85 ... (1)

    II condition

    x%=y%+0.5%

    x=y+0.5

    x-y=0.5 ... (2)

    solve (1) & (2)

    20 x - 10 y = 85 ... 1

    Total value

    1 x - 1 y = 0.50 ... 2

    Eliminate y

    multiply (1) by 1

    Multiply (2) by - 10

    20.00 x - 10.00 y = 85.00

    -10.00 x + 10.00 y = - 5.00

    Add the two equations

    10.00 x = 80.00

    / 10.00

    x = 8.00

    plug value of x in (1)

    20.00 x - 10.00 y = 85.00

    160.00 - 10.00 y = 85.00

    -10.00 y = 85.00 - 160.00

    -10.00 y = - 75.00

    y = 7.50

    x = 8.00 Interest rate on $14000

    y = 7.50 Interest rate on $7000
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