Solve the following system of equations

$$
\begin{aligned}
\left(4 t^{2}+t+4\right) x+(5 t+1) y & =4 t^{2}-t-3 \ldots \\
(t+2) x+\quad 2 y & =t \ldots
\end{aligned}
$$

Examine the signs of the roots for different values of $t$!

For which values of $t$ is the system of equations determined, indeterminate, or contradictory?
From equation 2) we have $y=\frac{t-(t+2) x}{2}$.

Substituting this expression for $y$ into 1) yields:

$$
2\left(4 t^{2}+t+4\right) x+(5 t+1) \cdot[t(t+2) x]=2\left(4 t^{2}-t-3\right) .
$$

After performing the indicated operations and simplifying:

$$
x=\frac{t^{2}-t-2}{t^{2}-3 t+2}=\frac{(t-2)(t+1)}{(t-2)(t-1)}=\frac{t+1}{t-1},
$$

provided that $t \neq 2$.

The system of equations is determined if $t^{2}-3 t+2 \neq 0$, that is, if $t \neq 2$ and $t \neq 1$. The corresponding value of $y$ is:

$$
y=\frac{t(t-1)-(t+2)(t+1)}{2(t-1)}=\frac{2 t+1}{t-1}
$$

The numerators and denominators in the expressions for $x$ and $y$ change sign at $t=-1,-\frac{1}{2},+1$.

For $t0$.

If $\quad 1<t$,
when $t=2$, the value of $x$ is of the form $\frac{0}{0}$, which is indeterminate; in this case, both equations can be reduced to the form

$$
2 x+y=1
$$

making the system indeterminate.

When $t=1$, the equations take the form

$$
9 x+6 y=0 \quad \text { and } \quad 3 x+2 y=1
$$

It is clear that we have a system of contradictory equations.