If x varies inversely as y and directly as t, and x =12 when t =10 and y =25, how do you find y when x is 6 and t= 3?

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Answer 1 · 2017-10-18T13:43:21

This is the only way I could think of that made the question work.
Someone else may differ on this.

$y=15$

$x$ varies inversely as $ycolor(white)("d") ->color(white)("dd") x=k/y" "..Eqn(1)$

and directly as t $color(white)("ddddd")->color(white)("dd")x=ct" "..Eqn(2)$

Given that $x=12$ when $t=10 and y=25$

Lets try combining these.

Consider $x=k/y$

Could the question be stating that the $k$ represents the 'varies directly part in which case we would have:

$x=(ct)/y larr" which combines "Eqn(1) and Eqn(2)$

Thus for initial condition we have: $c=(xy)/t = (12xx25)/10 = 30$

Consequently we have:

$x=(30t)/y color(white)("dddd")=>color(white)("dddd")y=(30t)/x$

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Find $y$ when $x=6 and t=3$

$=>y=(30xx3)/6=15$