5. Find all values of the parameter $a$ for which the system

$$
\left\{\begin{array}{l}
\frac{a^{2} x+2 a}{a x-2+a^{2}} \geqslant 0 \\
a x+a>\frac{5}{4}
\end{array}\right.
$$

has no solutions.
Answer: $a \in(-\infty ;-0.5] \cup\{0\}$

Solution. The value $a=0$ satisfies the condition of the problem. For $a \neq 0$, we make the substitution $t=a x$. For each $t$, there is a unique $x=\frac{t}{a}$.

$$
\left\{\begin{array} { l } 
{ \frac { a t + 2 a } { t - 2 + a ^ { 2 } } \geqslant 0 , } \\
{ t + a > \frac { 5 } { 4 } }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
\frac{a(t+2)}{t-2+a^{2}} \geqslant 0 \\
t+a>\frac{5}{4}
\end{array}\right.\right.
$$

We solve graphically on the Ota plane. The shaded area in the figure is the solution to the system. The coordinates of point A are found by solving the equation $a^{2}-a-\frac{3}{4}=0$.

![](https://cdn.mathpix.com/cropped/2024_05_06_55568ab07d9541f503efg-2.jpg?height=925&width=851&top_left_y=1602&top_left_x=571)