## Task 3 - 220913

a) Determine all real numbers $x$ for which the term $\frac{4 x-4}{2 x-3}$ is defined.

b) Among the numbers $x$ found in a), determine all those for which $0<\frac{4 x-4}{2 x-3}<1$ holds!
a) The term is undefined precisely when $2 x-3=0$. This is equivalent to $x=\frac{3}{2}$. Therefore, the term is defined for all real $x$ with $x \neq \frac{3}{2}$.

In the case $x>\frac{3}{2}$, we have $2 x-3>0$. Hence, multiplying an inequality by $2 x-3$ results in an equivalent inequality. Thus, the inequalities

$$
01$ and $x>\frac{3}{2}$, the inequality required in b) is satisfied.

In the case $x<\frac{3}{2}$, we have $2 x-3<0$. Therefore, multiplying an inequality by $2 x-3$ reverses the inequality. Thus, the inequalities

$$
0<\frac{4 x-4}{2 x-3} \quad \text{and} \quad 4 x-4<0 \quad \text{and} \quad 4 x-4>2 x-3 \quad \text{which means} \quad x<1 \quad \text{and} \quad x>\frac{1}{2}
$$

These two inequalities are satisfied precisely by all $x$ with $\frac{1}{2}<x<1$. For all these $x$, we also have $x<\frac{3}{2}$.

Thus, it is proven: The inequality required in b) holds exactly for all real $x$ with $\frac{1}{2}<x<1$.