Begin by making $t$ the subject in the $x$ equation. Multiply both sides by $t2$: $x(t2)=t$, and expand $xt2x=t$. Get the $t$ terms on one side: $xtt=2x$ and factorise $t(x1)=2x$. Hence $t=\frac{2x}{x1}$. Substituting into the $y$ equation gives $y=\frac{1}{\frac{2x}{x1}+1}$. Multiplying the top and bottom of this by $x1$ gives $y=\frac{x1}{2x+x1}$. Hence, the Cartesian equation is $y=\frac{x1}{3x1}$.
Parametric Equations – Cartesian Equation & Sketching
Parametric Equations are equations that are given in terms of a parameter. For example, $x =1/t$, $y=2t$ are parametric equations where $t$ is the parameter. You might notice that the word ‘parametric‘ is the adjective for ‘parameter‘. We often restrict the parameter, for example $0< t\leq 5$. See more on this example in the Curve Sketching section below.
In general, we write parametric equations in the form
$x=p(t)$, $y=q(t)$, for $t_1\leq t\leq t_2$.
In the general case, we may restrict $t$ to be between values $t_1$ and $t_2$. When $t$ represents time, we can use parametric equations to model motion. See Example 3.
If the question gives you $y$ in terms of $x$, this is a Cartesian Equation (named after Rene Descartes). Specifically, it is an explicit Cartesian equation because it is of the form $y=$ something. If the $x$s and $y$s are all mixed together in an equation, then it is an implicit Cartesian equation. A question may ask you to turn parametric equations in a Cartesian one. See below and Examples 1 and 2.
Cartesian Equation from Parametric Equations
Consider the equations above $x=1/t$, $y=2t$ for $0< t\leq 5$. We can find the Cartesian equation by eliminating $t$. We rearrange the $x$ equation to get $t=\frac{1}{x}$ and substituting gives $y=\frac{2}{x}$. Note that the $t$ values are limited and so will the $x$ and $y$ values be in the Cartesian equation. Note that this Cartesian equation is explicit. See Example 1 for another example of this.
It is not uncommon for a question to give the parametric equations in terms of trigonometric functions. For example, $x=\cos(\theta)$ and $y=\sin(\theta)$ for $0\leq\theta <2\pi$. Here, the parameter is $\theta$. In these cases, you should use trigonometric identities to obtain an implicit Cartesian equation. Note that in this example, $x^2+y^2=\cos^2\theta+\sin^2\theta=1$ and so $x^2+y^2=1$ is the implicit Cartesian equation. Recall that this is the unit circle with centre at the origin (more on circles). See Example 2 for another example using trigonometric identities.
Curve Sketching and Intersection Points
Consider again the example above: $x=1/t$, $y=2t$ for $0< t\leq 5$. We could sketch the graph given by these parametric equations in two ways. Firstly, we could find the Cartesian equation and sketch $y$ against $x$ for the restricted values. Or, secondly, it might be more convenient to calculate some values of $x$ and $y$ for some chosen values of $t$ in the given range. We can then plot $y$ against $x$ using the coordinate pairs.
t  1  2  3  4  5 
x  1  $\frac{1}{2}$  $\frac{1}{3}$  $\frac{1}{4}$  $\frac{1}{5}$ 
y  2  4  6  8  10 
Note that this is precisely the graph of $y=\frac{2}{x}$ for the restricted values of $x$ and $y$. Note also that $t$ cannot be zero but can get infinitely close to it. Hence, we include the part of the graph where $x\rightarrow\infty$ and $y\rightarrow 0$. See Example 2 for an example where we sketch by finding the Cartesian equation first. As we did in the example above, it is very important when sketching the curve of parametric equations to consider the domain and range properly.
Domain & Range
The $t$values in parametric equations are usually restricted. It follows that the $x$ and $y$ values will also be restricted. Recall that the domain of a function given explicitly ($y=f(x)$) are the values of $x$ that we can enter into the function. Also, the range of the function are the $y$values that can come out. See more on domain and range.
Consider the Cartesian equation $y=f(x)$ for the parametric equations $x=p(t)$ and $y=q(t)$ for $t_1\leq t\leq t_2$:
 the DOMAIN of $y=f(x)$ is the range of $p(t)$ and
 the RANGE of $y=f(x)$ is the range of $q(t)$
for the given $t$ values. See Example 3.
Intersection Points
A question might ask you to use the curve, or otherwise, to find various intersection points. These questions usually involve a certain amount of problem solving. For example, consider again the trigonometric parametric equations $x=\cos(\theta)$ and $y=\sin(\theta)$ for $0\leq\theta <2\pi$. The question might ask you to find the coordinates of where the curve crosses the $x$axis. Since we know this is the unit circle, the solutions are $(1,0)$ and $(1,0)$. If we didn’t have the curve we could solve $y=\sin(\theta)=0$ which is $\theta=0$ and $\theta=\pi$ on the given interval. Substituting these values of $\theta$ into $x$ we get $\cos(0)=1$ and $\cos(\theta)=1$. Hence, the coordinates are $(1,0)$ and $(1,0)$, as before. See Example 3 for a more complicated example of this.
Examples of Parametric Equations
Find the Cartesian equation for the parametric equations $x=\frac{t}{t2}$ and $y=\frac{1}{t+1}$ for $t>2$.
The parametric equations of a curve are $x=32\sin(\theta)$ and $y=2\cos(\theta)1$ for $0\leq\theta\leq\pi $ where $\theta$ is measured in radians anticlockwise from a line parallel to the $x$axis. Find the Cartesian equation and sketch the curve.
We can rewrite the equations as $x3=2\sin(\theta)$ and $y+1=2\cos(\theta)$. Squaring these gives $(x3)^2=4\sin^2(\theta)$ and $(y+1)^2=4\cos^2(\theta)$. Note that $4\sin^2(\theta)+4\cos^2(\theta)=4\left(\sin^2(\theta)+\cos^2(\theta)\right)=4$ (more on trigonometric identities). Hence, $(x3)^2+(y+1)^2=4$. This is a circle centred at $(3,1)$ with radius 2 (more on circles). However, since $\theta$ is restricted from 0 to $\pi$, this is the semicircle as follows:
A ball is thrown from the top of a cliff and the coordinates of its trajectory are given by the parametric equations $x=2t$ and $y=16+4t2t^2$ where $t$ is time, in seconds, $0\leq t\leq 4$.
 By first completing a table of coordinates, plot the projectile of the ball.
 State the domain and range of the parametric equations.
 Find the times at which the $y$coordinate of the ball is 17.

A typical table of all coordinates is given by:
t 0 1 2 3 4 x 0 2 4 6 8 y 16 18 16 10 0 Hence, we can plot the $x$ and $y$ coordinates to get the curve:
 From the curve, we can. see that the domain of the parametric equations is $0\leq x\leq 8$ and the range is $0\leq y\leq 18$.

The $y$coordinate is 17 when $16+4t2t^2=17$ or $2t^24t+1=0$. Completing the square gives $2(t1)^21=0$. Solving gives $t=1\pm\sqrt{\frac{1}{2}}$ or $t=1\pm\frac{\sqrt{2}}{2}$. To 3 decimal places, the solutions are $t=0.293$, and $t=1.707$. (Note that this gives $x=0.586$ and $x=3.414$ – symmetry across the line $x=2$ as expected).