If t^2-9/(t-3)^2 = s and s ≤ 3, what is the value of t in terms of s?

Question

If  \frac{t^2-9}{(t-3)^2}=s  and  s ≤ 3 , what is the value of t in terms of s?

(A)  \frac{(3 + s)}{(3 - s)}

(B)  \frac{9}{(s - 1)}

(C)  \frac{(s + 3)}{(s - 3)}

(D)  \frac{3(1 + s)}{(1 - s)}

(E)  \frac{(3 + 3s)}{(s - 1)}

amoghhlgr · Answer

Assume t=6.

(t+3)(t-3)/(t-3)^2 = (t+3)/(t-3) = 3

And s <= 3

For ‘t’ to be equal to 6, ‘s’ has to be equal to 2.

t = 3s is what I can think of…

Now plug in s=2 in all options and see which once equals to 6.

You can try this for other values of t.

So I guess it’s Option E

Posted from my mobile device

stne · Answer

Thank you for fixing the typo:

Given   =\frac{t^2-9}{(t-3)^2}=s

=\frac{t^2-3{^2}}{(t-3)^2} = \frac{(t+3) (t-3)}{(t-3)^2} =s

=\frac{t+3}{t-3} =s

t+3 = s(t-3)

t+3 =st-3s

3+3s =st-t

3+3s =t(s-1)

t= \frac{3+3s}{s-1}

Ans E

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