## Precalculus (6th Edition) Blitzer

The different sets of parametric equation are as follows: $x=t\text{, }\,y=4t-3\,\,\text{ and }\,\,x=t+1\text{,}\,\,y=4t+1$
Choose x values such that solution set for y doesn’t change. Let us assume $x=t$, then $y=4t-3$ Here, for $x=t$ And the solution set for y remains the same for both negative and positive values. Again, let $x=\,{{t}^{2}}$; then $y=4{{t}^{2}}-3$ The solution set for y is positive for both negative and positive values of t. Therefore, $x=\,{{t}^{2}}$ It cannot be used to determine the set of parametric equations. Again, Let $x=\,t+1$, then \begin{align} & y=4\left( t+1 \right)-3 \\ & y=4t+4-3 \\ & y=4t+1 \\ \end{align} The solution set for y remains the same for both negative and positive values. Note that many different sets of parametric equations are possible. Choose any two from them. Thus, $x=\,t$, $y=4t-3$ and $x=\,t+1$, $y=4t+1$ are two different sets of parametric equations.