Parametric equations, eliminating the parameter $\,x = t^2 + t,\,$ $y= 2t-1$ $$x = t^2 + t\qquad y= 2t-1$$
So I solve $y$ for $t$
$$t = \frac{1}{2}(y+1)$$
Then I am supposed to plug it into the equation of $x$ which is where I lose track of the logic.
$$x = \left( \frac{1}{2}(y+1)\right)^2 +  \frac{1}{2}(y+1) = \frac{1}{4}y^2 + y+\frac{3}{4}$$
That is now my answer? I am lost. This is x(y)? How is this valid? I don't understand.

Let's assume you are walking on an xy-plane. Your x-position (or east-west position) at a certain time t is given by $x = t^2 + t$, and your y-position is $y = 2t - 1$.
If you want to know what the whole path you remained is, without wanting to know when you stepped on where? Eliminate t:
$$x = \frac{1}{4} y^2 + y + \frac{3}{4}$$
If a pair of $(x, y)$ satisfied that, that means some time in the past or in the future, you stepped or would step on that point.
