Given \(\displaystyle x=3t+7\) and \(\displaystyle y=7t+3\), let's solve both equations for \(\displaystyle t\):

\(\displaystyle x=3t+7\rightarrow t=\frac{x-7}{3}\)

\(\displaystyle y=7t+3\rightarrow t=\frac{y-3}{7}\)

Since both equations equal \(\displaystyle t\), let's set them equal to each other and solve for \(\displaystyle y\):

\(\displaystyle \frac{y-3}{7}=\frac{x-7}{3}\)

\(\displaystyle y-3=7(\frac{x-7}{3})\)

\(\displaystyle y=\frac{7}{3}(x-7)+3\)

Given \(\displaystyle x=7t+5\) and \(\displaystyle y=11t+2\), let's solve both equations for \(\displaystyle t\):

\(\displaystyle x=7t+5\rightarrow t=\frac{x-5}{7}\)

\(\displaystyle y=11t+2\rightarrow t=\frac{y-2}{11}\)

Since both equations equal \(\displaystyle t\), let's set them equal to each other and solve for \(\displaystyle y\):

\(\displaystyle \frac{y-2}{11}=\frac{x-5}{7}\)

\(\displaystyle y-2=11(\frac{x-5}{7})\)

\(\displaystyle y=\frac{11}{7}(x-5)+2\)

Given \(\displaystyle x=-6t+11\) and \(\displaystyle y=2t-10\) let's solve both equations for \(\displaystyle t\):

\(\displaystyle x=-6t+11\rightarrow t=-\frac{x-11}{6}\)

\(\displaystyle y=2t-10\rightarrow t=\frac{y+10}{2}\)

Since both equations equal \(\displaystyle t\), let's set them equal to each other and solve for \(\displaystyle y\):

\(\displaystyle \frac{y+10}{2}=-\frac{x-11}{6}\)

\(\displaystyle y+10=-2(\frac{x-11}{6})\)

\(\displaystyle y=-\frac{x-11}{3}-10\)

To convert the equation from parametric form to rectangular form, we must eliminate the parameter. To do this, we can solve for t with respect to x:

\(\displaystyle t=\pm \sqrt{\frac{x}{2}}\)

The plus or minus is important to remember because a square root was taken.

Now, simply plug this into the equation for y:

\(\displaystyle y=\pm10\sqrt{\frac{x}{2}}\)

To convert a parametric equation into a rectangular equation, we must eliminate the parameter. We were already given an equation where t was in terms of just a variable:

\(\displaystyle t=y\)

Next, substitute this into the equation for x which contains t:

\(\displaystyle x=2(y^3)+5\)

Finally, rearrage and solve for y:

\(\displaystyle y=\sqrt[3]{\left(\frac{x-5}{2}\right)}\)

To convert the given parametric equation into rectangualr coordinates, we must eliminate the parameter by solving for t:

\(\displaystyle t=\arccos (x)\)

Now replace t with the new term in the equation for y:

\(\displaystyle y= (\cos(\textup{arccos}(x)^2)+3=x^2+3\)

To convert from parametric to rectangular form, eliminate the parameter (t) from one of the equations:

\(\displaystyle t=\ln(x)\)

Now plug this into the equation for y to get our final answer:

\(\displaystyle y=2\sin(\ln(x))\)

Given \(\displaystyle x=3t-6\) and \(\displaystyle y=7t+9\), let's solve both equations for \(\displaystyle t\):

\(\displaystyle x=3t-6\rightarrow t=\frac{x+6}{3}\)

\(\displaystyle y=7t+9\rightarrow t=\frac{y-9}{7}\)

Since both equations equal \(\displaystyle t\), let's set them equal to each other and solve for \(\displaystyle y\):

\(\displaystyle \frac{y-9}{7}=\frac{x+6}{3}\)

\(\displaystyle y-9=\frac{7}{3}(x+6)\)

\(\displaystyle y-9=\frac{7}{3}x+14\)

\(\displaystyle y=\frac{7}{3}x+23\)

Given \(\displaystyle x=2t+11\) and \(\displaystyle y=11t+2\),  let's solve both equations for \(\displaystyle t\):

\(\displaystyle x=2t+11\rightarrow t=\frac{x-11}{2}\)

\(\displaystyle y=11t+2\rightarrow t=\frac{y-2}{11}\)

Since both equations equal \(\displaystyle t\), let's set them equal to each other and solve for \(\displaystyle y\):

\(\displaystyle \frac{y-2}{11}=\frac{x-11}{2}\)

\(\displaystyle y-2=\frac{11}{2}(x-11)\)

\(\displaystyle y=\frac{11}{2}(x-11)+2\)

Given \(\displaystyle x=t+3\) and \(\displaystyle y=\frac{1}{2}t-3\),  let's solve both equations for \(\displaystyle t\):

\(\displaystyle x=t+3\rightarrow t=x-3\)

\(\displaystyle y=\frac{1}{2}t-3\rightarrow t=2(y+3)\)

Since both equations equal \(\displaystyle t\), let's set them equal to each other and solve for \(\displaystyle y\):

\(\displaystyle 2(y+3)=x-3\)

\(\displaystyle y+3=\frac{x-3}{2}\)

\(\displaystyle y=\frac{x-3}{2}-3\)