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So I have:

$$y=\frac{t+2}{t-3},\qquad x=\frac{3t+1}{t-4}$$

What is $\dfrac{\mathrm dy}{\mathrm dx}$ when $t=1$?

I got $\dfrac{45}{52}$ but wanted to check the answer.

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$dy/dx=(dy/dt)/(dx/dt)=(-5/(x-3)^2)/(-13/(x-4)^2)$ By substituting $t=1$, you can get $dy/dx=(-5/4)/(-13/9)=45/52$

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