## Task B-1.6.

Determine the value of the real parameter $a$ for which each of the equations

$$
2 a-1=\frac{3-3 a}{x-1} \quad \text { and } \quad a^{2}(2 x-4)-1=a(4-5 x)-2 x
$$

has a unique solution and their solutions are equal.
## Solution.

First, let's solve the equation $2 a-1=\frac{3-3 a}{x-1}$. Note that $x \neq 1$ must hold. After multiplying the equation by $x-1$ and simplifying the resulting algebraic expressions, we get:

$$
\begin{gathered}
2 a x-2 a-x+1=3-3 a \\
x(2 a-1)=2-a \\
x=\frac{2-a}{2 a-1}, a \neq \frac{1}{2}
\end{gathered}
$$

Since $x \neq 1$, from $\frac{2-a}{2 a-1} \neq 1$ it follows that

$$
2-a \neq 2 a-1,
$$

which means $a \neq 1$.

Thus, for $a \in \mathbf{R}, a \neq \frac{1}{2}$ and $a \neq 1$, the equation has a unique solution $x=\frac{2-a}{2 a-1}$.

Now, let's solve the second equation.

$$
\begin{gathered}
a^{2}(2 x-4)-1=a(4-5 x)-2 x \\
2 x a^{2}-4 a^{2}-1=4 a-5 a x-2 x \\
x\left(2 a^{2}+5 a+2\right)=4 a^{2}+4 a+1 \\
x\left(2 a^{2}+4 a+a+2\right)=(2 a+1)^{2} \\
x[2 a(a+2)+a+2]=(2 a+1)^{2} \\
x(a+2)(2 a+1)=(2 a+1)^{2}
\end{gathered}
$$

From this equality, we conclude that for $a \in \mathbf{R}, a \neq-\frac{1}{2}$ and $a \neq-2$,

the equation has a unique solution

$$
x=\frac{2 a+1}{a+2} \text {. }
$$

Let's equate the solutions of the given equations.

$$
\frac{2-a}{2 a-1}=\frac{2 a+1}{a+2}
$$

This leads to $4-a^{2}=4 a^{2}-1$, so $a^{2}=1$, and

$$
a= \pm 1
$$

Since $a \neq 1$, it follows that the equations have the same solution only for $a=-1$.

Note: Instead of calculating the value of $a$ for which the condition $x \neq 1$ is satisfied, the student can check at the end whether the given equations have the same solution for the obtained real numbers $a= \pm 1$. In this case, the student can get all the expected points, but must have written the conditions for which these equations have a unique solution, i.e., $a \neq \frac{1}{2}$ for the first equation and $a \neq-\frac{1}{2}, a \neq-2$ for the second equation.

If the student has not written only these conditions, deduct 2 points. If the student has not written the condition $x \neq 1$, and has not discarded one solution and has not even checked the solutions, but has the previously mentioned conditions for the solutions of the equations individually, deduct 3 points.

If the student has none of the conditions, no checks, only the solution, they can get a maximum of 5 points.