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#