How to find the Cartesian equation of a plane curve from a parametric equation? More specifically, how to express 
$$\begin{aligned}x(t) &=\frac{2t}{1+t^2}\\
y(t) &=\frac{1-t^2}{1+t^2}\end{aligned}$$
in terms of $x$ and $y$? I attempted adding the two, getting a square from the numerator and a few other methods, but running out of time.
EDIT: please don't work backwards, and try to show this as simple as possible. Overcomplication or overdefining equations may confuse me even more.

Let define the following polynomials: $$\begin{align}P_x(t)&:=xt^2-2t+x,\\Q_y(t)&:=(y+1)t^2+y-1.\end{align}$$
Let $\mathcal{C}$ be the curve defined by the given parametric equations, notice that: $$(x,y)\in\mathcal{C}\Leftrightarrow\exists t\textrm{ s.t }P_x(t)=0=Q_y(t).$$
Furthermore, notice that if $(x,y)\neq (0,-1)$, $\deg(P_x)=2=\deg(Q_y)$. Therefore, one has: $$(x,y)\in\mathcal{C}\setminus\{(0,-1)\}\Leftrightarrow\textrm{Res}_{2,2}(P_x,Q_y)=0.$$
Where $\textrm{Res}_{2,2}(P_x,Q_y)$ is the resultant in degree $(2,2)$ of $P_x$ and $Q_y$, when computing it one gets: $$\textrm{Res}_{2,2}(P_x,Q_y)=4(x^2+y^2-1).$$
Whence, $\mathcal{C}\setminus\{(0,-1)\}$ is given by $4(x^2+y^2-1)=0$, that is $x^2+y^2=1.$ Finally, notice that $(0,-1)$ satisfies the same equation to conclude that: $$\mathcal{C}:x^2+y^2=1.$$
Remark. This method is universal to all unicursal plane curves, that is plane curves given by a rational parametrization.
