3. Determine all real numbers $x$ that satisfy the equation $4 x-2\lfloor x\rfloor=5$. (The symbol $\lfloor x\rfloor$ denotes the greatest integer not greater than the number $x$, i.e., the floor of the real number $x$.)
SOLUTION. Let $\lfloor x\rfloor=a$, then $x=a+t$, where $t \in [0,1)$, and the equation $4(a+t)-2a=5$ can be equivalently rearranged to the form $a=\frac{5}{2}-2t$. For the number $a$ to be an integer, $2t$ must be $k \cdot \frac{1}{2}$, where $k$ is an odd number. Additionally, $2t \in [0,2)$. Therefore, either $2t=\frac{1}{2}$ and $a=2$, or $2t=\frac{3}{2}$ and $a=1$. The original equation thus has two solutions: $x_1=2.25$ and $x_2=1.75$.

ANOTHER SOLUTION. We rearrange the equation to the form $2x-\frac{5}{2}=\lfloor x\rfloor$. Its solutions are the $x$-coordinates of the intersection points of the graphs of the functions $l: y=2x-\frac{5}{2}$ and $p: y=\lfloor x\rfloor$. The graphs intersect at two points, as we can see in Fig. 3. For the first intersection point, $\lfloor x\rfloor=1$. Substituting into the original equation, we get $4x-2=5$ and thus $x_1=\frac{7}{4}=1.75$. For the second intersection point, $\lfloor x\rfloor=2$, so $4x-4=5$ and $x_2=\frac{9}{4}=2.25$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5e5d61e88a21dd6b9af7g-4.jpg?height=745&width=851&top_left_y=1278&top_left_x=611)

Fig. 3