How many Eggs??

Let the original number of eggs be $n$ .

• To the first customer he sold $\dfrac n2 + \dfrac 12$ and number of eggs remained was $n - \dfrac n2 - \dfrac 12 = \dfrac n2 - \dfrac 12$ .
• To the second customer he sold $\dfrac 12 \left(\dfrac n2 - \dfrac 12\right) + \dfrac 12$ and number of eggs remained was $\dfrac n2 - \dfrac 12 - \dfrac 12 \left(\dfrac n2 - \dfrac 12\right) - \dfrac 12 = \dfrac n4 - \dfrac 34$ .
• To the third customer he sold $\dfrac 12 \left(\dfrac n4 - \dfrac 34\right) + \dfrac 12$ and number of eggs remained was $0$ .

Then, we have:

$\begin{aligned} n & = \frac n2 + \frac 12 + \frac 12 \left(\frac n2 - \frac 12\right) + \frac 12 + \frac 12 \left(\frac n4 - \frac 34\right) + \frac 12 & \small \blue {\text{Multiply both sides by 8.}} \\ 8n & = 4n+4 + 2n - 2 + 4 + n - 3 + 4 \\ \implies n & = \boxed 7 \end{aligned}$

Going backwards: if $z$ is the number of eggs the man had before he sold eggs to the third customer, then $z = z/2 + 1/2$ . Thus $z = 1$ .

If $y$ is the number of eggs the man had before he sold eggs to the second customer, then $y = y/2 + 1/2 + 1$ . Thus $y = 3$ .

If $x$ is the number of eggs the man had before he sold eggs to the first customer, then $x = x/2 + 1/2 + 3$ . Thus $x = 7$ .

p.s. It was unnecessary to mention that the man had not broken any eggs. The values of $x,y$ , and $z$ are integral from the algebra. There is no non-integer solution to this system.

Solving the equation $x=\dfrac{x}{2}+\dfrac{1}{2}+\dfrac{x-1}{4}+\dfrac{1}{2}+\dfrac{x-3}{8}+\dfrac{1}{2}$ for $x$ we get the answer as $x=7$