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17 July, 09:52

if the value of x varies directly as the value of y, and x = 16 when y = 3/4, what is the value of x when y = 3/8?

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  1. 17 July, 12:21
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    x=8

    Step-by-step explanation:

    If the value of x varies directly as the value of y we can say they are directly proportional. This implies that when y increases x will increase too, and when y decreases x will also decrease.

    If x=16 when y=3/4 we can write their relation as:

    x/y = 16 / (3/4) = 16 * (4/3) = 64/3

    So, for every value of y, the ratio x/y must maintain equal to 64/3 because both values are proportional.

    Now we can see what happens to x when y varies. If y=3/8:

    x / (3/8) = 64/3

    Multiply both sides by 3/8 to eliminate the 3/8 in the denominator.

    x / (3/8) * (3/8) = (64/3) * (3/8)

    See that we have a 3 multiplying and another dividing, we can eliminate them.

    x = 64/8

    x = 8

    So, if y decreases to 3/8 x will decrease to 8. Notice that y is decreasing in a half as 3/8 is the half of 3/4, and x is also decreasing in a half from 16 to 8, what makes sense.
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