Solve (x-1)(x-z)geq0

$$(x-1)(x-z) \geq 0$$

Solve for x

$$\left\{\begin{matrix}\\x=1\text{; }x=z\text{, }&\text{unconditionally}\\x\in (-\infty,z]\cup [1,\infty)\text{, }&z<1\\x\in (-\infty,1]\cup [z,\infty)\text{, }&z\geq 1\end{matrix}\right.$$

Solution Steps

For the product to be ≥0, x-1 and x-z have to be both ≤0 or both ≥0. Consider the case when x-1 and x-z are both ≤0.

$$x-1\leq 0 x-z\leq 0$$

The solution satisfying both inequalities is \left\{\begin{matrix}x\leq z\text{, }&z\leq 1\\x\leq 1\text{, }&z>1\end{matrix}\right..

$$\left\{\begin{matrix}x\leq z\text{, }&z\leq 1\\x\leq 1\text{, }&z>1\end{matrix}\right.$$

Consider the case when x-1 and x-z are both ≥0.

$$x-z\geq 0 x-1\geq 0$$

The solution satisfying both inequalities is \left\{\begin{matrix}x\geq 1\text{, }&z\leq 1\\x\geq z\text{, }&z>1\end{matrix}\right..

$$\left\{\begin{matrix}x\geq 1\text{, }&z\leq 1\\x\geq z\text{, }&z>1\end{matrix}\right.$$

The final solution is the union of the obtained solutions.

$$\left\{\begin{matrix}x\leq z\text{, }&z\leq 1\\x\leq 1\text{, }&z>1\\x\geq z\text{, }&z\geq 1\\x\geq 1\text{, }&z<1\end{matrix}\right.$$

Solve for z

$$\left\{\begin{matrix}\\z=x\text{, }&\text{unconditionally}\\z\geq x\text{, }&x\leq 1\\z\leq x\text{, }&x\geq 1\\z\in \mathrm{R}\text{, }&x=1\end{matrix}\right.$$