Solve y_{1}(t)=8tquady_{2}(t)=-8t+54

$$y _ { 1 } ( t ) = 8 t \quad y _ { 2 } ( t ) = - 8 t + 54$$

Solve for y_2

$$y_{2}=\frac{27-4t}{4t^{2}}$$

$$t\neq 0$$

Steps for Solving Linear Equation

Multiply t and t to get t^{2}.

$$8t^{2}y_{2}=-8t+54$$

The equation is in standard form.

$$8t^{2}y_{2}=54-8t$$

Divide both sides by 8t^{2}.

$$\frac{8t^{2}y_{2}}{8t^{2}}=\frac{54-8t}{8t^{2}}$$

Dividing by 8t^{2} undoes the multiplication by 8t^{2}.

$$y_{2}=\frac{54-8t}{8t^{2}}$$

Divide -8t+54 by 8t^{2}.

$$y_{2}=\frac{\frac{27}{4}-t}{t^{2}}$$

Solve for t (complex solution)

$$\left\{\begin{matrix}t=\frac{\sqrt{27y_{2}+1}-1}{2y_{2}}\text{; }t=-\frac{\sqrt{27y_{2}+1}+1}{2y_{2}}\text{, }&y_{2}\neq 0\\t=\frac{27}{4}=6.75\text{, }&y_{2}=0\end{matrix}\right.$$

Solve for t

$$\left\{\begin{matrix}t=\frac{\sqrt{27y_{2}+1}-1}{2y_{2}}\text{; }t=-\frac{\sqrt{27y_{2}+1}+1}{2y_{2}}\text{, }&y_{2}\neq 0\text{ and }y_{2}\geq -\frac{1}{27}\\t=\frac{27}{4}=6.75\text{, }&y_{2}=0\end{matrix}\right.$$