Given that
[tex]\begin{gathered} x(t)=2t-1 \\ y(t)=5t \end{gathered}[/tex]In both cases, we will make t the subject of the formula and equate both equations together
[tex]\begin{gathered} x=2t-1 \\ 2t=x+1 \\ \frac{2t}{2}=\frac{x+1}{2} \\ t=\frac{x+1}{2}\ldots\ldots\ldots\ldots\ldots(1) \end{gathered}[/tex][tex]\begin{gathered} y=5t \\ t=\frac{y}{5}\ldots\ldots\ldots\ldots\ldots(2) \end{gathered}[/tex]To get the cartesian form, equate equation (1) and (2) together
[tex]\begin{gathered} \frac{x+1_{}}{2}=\frac{y}{5} \\ by\text{ cross multiplying, we will have} \\ 2y=5(x+1)_{} \\ 2y=5x+5 \\ \text{divide all through by 2 we will have} \\ \frac{2y}{2}=\frac{5x}{2}+\frac{5}{2} \\ y=\frac{5}{2}x+\frac{5}{2} \end{gathered}[/tex]Hence,
The correct answer is OPTION C
