Parametric to Cartesian Assitance

Question

I can't seem to a way to convert this parametric equation into cartesian where:

$$x = \frac{1-t^2}{1+t^2}$$ $$y = \frac{2t}{1+t^2}$$

and $3y=4x$

Answer 1

$$x = \frac{1-t^2}{1+t^2} \implies x^2 = \frac{(1-t^2)^2}{(1+t^2)^2}$$

$$y = \frac{2t}{1+t^2} \implies y^2 = \frac{4t^2}{(1+t^2)^2} $$

$$ x^2+y^2 = \frac {(1-t^2)^2 +4t^2}{ (1+t^2)^2}=1$$ Thus we have $x^2+y^2 =1$, and $3y=4x$ Now it is straigth forward from here on.

Answer 2

I can't seem to a way to convert this parametric equation into cartesian where:

It is not clear what you mean by that.

$x = \dfrac{1-t^2}{1+t^2}\,,\; y = \dfrac{2t}{1+t^2}\,,\;$ and $\;3y=4x$

If you just want to solve the system, substitute $\,x,y\,$ from the first two equations into the third:

$$\require{cancel} 3y=4x \;\;\iff\;\; 3 \, \frac{1-t^2}{\cancel{1+t^2}} = 4 \frac{2t}{\cancel{1+t^2}} \;\;\iff\;\; 3 t^2 + 8 t - 3 = 0 \;\;\iff\;\; t = \ldots $$