“n” is inversely proportional to t+3. If t = 1 when n = 3, find t when n = 4

Here is my work:

n = k/t +3

3 = k/1+3

3(4) = k

12 = k

... find t when n = 4:

y = 12/n

y = 12/4 = 3/1 = 3

y = 3

Thanks(Clapping)

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- Aug 12th 2009, 03:11 AMbobbyboy1111Inverse variation problem (please check)
“n” is inversely proportional to t+3. If t = 1 when n = 3, find t when n = 4

Here is my work:

n = k/t +3

3 = k/1+3

3(4) = k

12 = k

... find t when n = 4:

y = 12/n

y = 12/4 = 3/1 = 3

y = 3

Thanks(Clapping) - Aug 12th 2009, 03:51 AMEphemerasylum
Nicely done for the first half of your solution,

but after you found k = 12 you made a tiny mistake.

$\displaystyle n = \frac{12}{t+3} $

In order to find t, we substitute n by applying the given condition that n = 4.

This operation gives

$\displaystyle 4 = \frac{12}{t+3} $

Then we put them in a better order

$\displaystyle t+3 = \frac{12}{4} = 3 $

At last we find t = 0.

So your work is basically okay but incomplete ~