So I got the tangent line of the first equation to be 12t/(3t^2+4) and I changed the second parametric equation to the cartesian form and got y= -(12/7x+5) with 12/7 as my slope. I equated 12t/(3t^2+4)= 12/7 and solve for t using the quadratic formula. I got the values -1 and -4/3 but they are incorrect. So, I'm confused where I went wrong. 
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user226550
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From $y = (-12/7)x - 5$, the slope of the second line should be -12/7 instead.
Also, the solutions to $\frac{12t}{3t^2+4} = \frac{12}{7}$ are $1, \frac{4}{3}$ and $-1, -\frac{4}{3}$ are instead the solutions to $\frac{12t}{3t^2+4} = -\frac{12}{7}$
For the points $(y(t), x(t))$ of the solution, you should plug in $t = -1, -4/3$ into your equation for $x,y$.
Otherwise , your general approach is fine.
Rolf Hoyer
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Hey, I thought the equation of a line was in the form y=mx+b, so wouldn't the equation be y= -(12/7x +5) where the slope is 12/7? And even if the slope is -12/7, the values of -1 and -4/3 for the x value of the points are still incorrect :/ so I'm a bit lost. – user226550 Apr 09 '15 at 02:17
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$y = -(mx+b)$ is equivalent to $y = -mx -b$, so the slope is negative. See my edited answer w.r.t. getting $x$-values. – Rolf Hoyer Apr 09 '15 at 02:20
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:/ I still get the values of -1 and -4/3 with the slope of -12/7? – user226550 Apr 09 '15 at 02:25
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remember that you have solved for $t$ values, and need to get $x$- and $y$-values for your answer – Rolf Hoyer Apr 09 '15 at 02:26
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ohhhhh damn you are right. :) thanks – user226550 Apr 09 '15 at 02:33