# Systems of inequations with one variable

A system of inequations with one variable is a set of inequations of a variable that act simultaneously, that is, the solution points must satisfy all the inequations of the system.

An example of a system of inequations is:

$$\left\{ \begin{array}{l} x-1 > 0 \\ x+2(1-x) < 4+x \\ 2x < 8 \end{array}\right.$$

In this example we can see that the respective solutions of every inequation are: $$\left\{ \begin{array}{l} x > 1 \\ x > -1 \\ x < 4 \end{array}\right.$$

and, as the solution of the system must satisfy every inequation, it is clear that it will be satisfied only by the points between $1$ and $4$ (the solution is $1 < x < 4$).

As we have already seen in the example, solving a system of inequations with one variable consists of resolving each system separately and in the end taking the intersection of the sets of solutions of each one; in other words, taking the most restrictive inequations.

The systems of inequations with one variable can be formed by first degree inequations and quadratic inequations (in fact, there can appear systems of inequations of any degree and not even polynomial ones, but the resolution of these is much more complicated). In these cases (first and second degree), we will solve every inequation and later we will take the total intersection of solutions.

It is possible that sometimes, when we are looking for the intersection of the solutions sets, we realize that this one could not exist since the inequations are incompatible (in such a case we will say that the system has no solution) or that the set comes down to a point.

For example:

• $x < 1$ and $x > 3$ This set of solutions of two inequations is incompatible, therefore we say that the system has no solution.

• $x \leqslant 5$ and $x \geqslant 5$ This set of solutions of two inequations is compatible, and the intersection of these is the number $x = 5$.

At last, let's take a look at an example of a system of inequations with one variable.

Let's consider the system: $$\left\{ \begin{array}{l} 2x-4 > 2 \\ 2x-1 < 7 \end{array}\right.$$

We are going to solve it by isolating $x$ in both inequations: $$\left\{ \begin{array}{l} 2x-4 > 2 \\ 2x-1 < 7 \end{array}\right. \Rightarrow \left\{ \begin{array}{l} x > \dfrac{2+4}{2}=3 \\ x < \dfrac{7+1}{2}=4 \end{array}\right.$$

therefore the solution is $3 < x < 4$.