For what values of $c$ do the following equations have solutions?

$$
\text { a) } x+\left[\frac{x}{2}\right]=c ; \quad \text { b) } x-\left[\frac{x}{2}\right]=c
$$

If there are solutions, how many solutions are there, and what are the solutions?
Graphically, we obtain the solution of the given equations by plotting the left side and intersecting it with the line $y=c$ representing the right side.

The $[x / 2]$ term on the left side is a step function, constant over intervals of 2 units, with jumps of 1 unit at even $x$ values. Therefore, the graph of the left side of our equations is obtained by interrupting the line $y=x$ at every integer abscissa value of $2k$ and shifting the segment between abscissas $2k$ and $2k+2$ (including the left endpoint but not the right) by $k$ units in the $y$ direction.

![](https://cdn.mathpix.com/cropped/2024_05_02_e249fcf36b102f744371g-1.jpg?height=477&width=330&top_left_y=412&top_left_x=884)

Figure 1

Thus, the image of the left side of equation (a) over the half-closed interval $[2k, 2k+2)$ on the $x$-axis has ordinates satisfying $3k \leq x + \left[\frac{x}{2}\right] < 3k + 2$, and over the next interval $[2k+2, 2k+4)$, the ordinates satisfy $3k + 3 \leq x + \left[\frac{x}{2}\right] < 3k + 5$. Therefore, the values $3k + 2 \leq y < 3k + 3$ do not belong to the range. If $c$ is such a value, i.e., if

$$
\left[c - 3\left[\frac{c}{3}\right]\right] = 2
$$

then equation (a) has no solution; otherwise, it has one solution. Specifically, if $3k \leq c < 3k + 2$, then, since the abscissas of the points on the graph segment are $k$ units smaller than their ordinates, the intersection point is given by

$$
x = c - k = c - \left[\frac{c}{3}\right]
$$

(Validity condition in another form:

$$
\left[c - 3\left[\frac{c}{3}\right]\right] = 0 \quad \text{or} \quad 1.
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_e249fcf36b102f744371g-1.jpg?height=485&width=575&top_left_y=1557&top_left_x=756)

Figure 2

For equation (b), similarly,

$$
\text{if } 2k \leq x < 2k + 2
$$

then $\quad k \leq x - \left[\frac{x}{2}\right] < k + 2$,

and if $2k + 2 \leq x < 2k + 4$,

$$
\text{then } \quad k + 1 \leq x - \left[\frac{x}{2}\right] < k + 3,
$$

which means the left side takes the values

$$
k + 1 \leq c < k + 2
$$

in both intervals. According to this, every real value is taken exactly twice, and the equation has two solutions for any $c$. The abscissa of the intersection point of the graph and the line $y=c$ is $k$ units larger in interval (1) and $k + 1$ units larger in interval (2) than the ordinate, and according to (3), $k = [c] - 1$, so the solutions are:

$$
x_1 = c + [c] - 1, \quad \text{or} \quad x_2 = c + [c].
$$