# First degree inequations

An inequation is an algebraic expression formed by numbers, a variable that we will call $x$ and a symbol of inequality.

Examples of inequations would be:

1. $x < 2$

2. $4x+2\geqslant -1$

3. $-x > -3+2x$

In these cases, we would say that the inequation 1 would already be solved, because if $x$ takes values less than $2$ it will always satisfy the inequality, and inequations 2 and 3 would be resolved, that is, and find what values of $x$ satisfy the respective inequations.

## Solution of an inequation

Given an inequation, we will think that we have solved it when we find an expression like $x < a$, $x > a$, $x\leqslant a$ or $x\geqslant a$, $a$ being a number. Once we have this expression we can already say that for the inequation to be true, $x$ should satisfy the condition found and we will claim that the inequation is solved.

Also, these should be examples of solutions: $$x < 2, \ x > 3, \ x\leqslant -1, \ x\geqslant 6$$

També serien exemples de solucions: $$2 < x, \ -1\geqslant x$$

since using the property of symmetry of the inequalites, they are equivalent to $$x > 2, \ x\leqslant -1$$

## Resolution of inequations

Likewise, we have learned to solve first degree equations, so now we are going to learn how to solve first degree inequations.

The method to solve these inequations is the same as that for solving equations, even though there are small changes.

To start, let's see the analogy that exists between solving a first degree equation and a first degree inequation:

We will solve the equation $2(x-5)=2$ and the inequation $2(x-5)\geqslant 2$.

Let's solve the equation: $$2(x-5)=2 \Rightarrow 2x-10=2 \Rightarrow 2x=2+10 \Rightarrow 2x=12 \Rightarrow x=\dfrac{12}{2} \Rightarrow x=6$$

and we say that the solution is $x=6$.

On the other hand, let's solve the inequation: $$2(x-5)\geqslant2 \Rightarrow 2x-10\geqslant2 \Rightarrow 2x\geqslant2+10 \Rightarrow 2x\geqslant12 \Rightarrow x\geqslant\dfrac{12}{2} \Rightarrow x\geqslant6$$

and we say that the solution is $x\geqslant6$, that is, $x$ can take any value greater than or equal to six.

Notice that the resolution method has been the same for both exercises, therefore: what is the difference between the process of resolution of an inequation and an equation?

To answer this question, let's see how the inequalites change when we operate with additions, subtractions, multiplications and divisions:

Let $A$, $B$ and $C$ be any three numbers, then:

if $A < B \Rightarrow \left\{ \begin{array}{l} A+C < B+C \\ A-C < B-C \end{array} \right.$

if $A > B \Rightarrow \left\{ \begin{array}{l} A+C > B+C \\ A-C > B-C \end{array} \right.$

As we can see, we can add or subtract the same value on each side of the inequality without having problems with the symbol of the inequality.

This property has already been studied in the topic on equations since we could add or subtract the same value on each side of the equality.

This property allows us to add and to subtract the same value on each side of an inequality of an inequation in order to be able to isolate the variable $x$ on one side of the inequation.

Let’s solve the inequation $x+3 < 4$. $$x+3 < 4 \Rightarrow x+3-3 < 4-3 \Rightarrow x < 4-3 \Rightarrow x < 1$$

## Multiplication and division

Multiplying and dividing by a value in an inequation, it is possible that we must change the symbol in the inequality: from less than to greater than or vice versa (the same goes for less than or equal to, and the other way round).

Let $A$, $B$ and $C$ be any three numbers, then:

• If $C$ is positive and $A < B$ then $A\cdot C < B\cdot C \$ and $\ \dfrac{A}{C} < \dfrac{B}{C}$ (the inequality does not change).

• If $C$ is positive and $A > B$ then $A\cdot C > B\cdot C \$ and $\ \dfrac{A}{C} > \dfrac{B}{C}$ (the inequality does not change).

• If $C$ is negative and $A < B$ then $A\cdot C > B\cdot C \$ and $\ \dfrac{A}{C} > \dfrac{B}{C}$ (the inequality does not change).

• If $C$ is negative and $A > B$ then $A\cdot C < B\cdot C \$ and $\ \dfrac{A}{C} < \dfrac{B}{C}$ (the inequality does not change).

The reason for the change in the order of the inequality if we multiply or divide by a negative number will be clearly seen in an example:

If $A = 2$ and $B = 3$ (we have that $A < B$ because $2 < 3$), then, we multiply by $(-1)$ and we obtain: $$\left. \begin{array}{l} 2\cdot (-1)=-2 \\ 3\cdot (-1) =-3 \end{array} \right\} \Rightarrow -2<-3 \ \text{ FALSE, } \ -2 > -3 \ \text{ TRUE}$$

We can see that we have had to change the order of the inequality in order to keep the expression true.

This property allow us to multiply and to divide by the same value on both sides of an inequation (in a similar way as we were doing with equations), and in this way we will be able to isolate our variable $x$ without problems on one of the sides of the inequation.

Given the inequation $3x < 6$, let's solve it: $$3x < 6 \Rightarrow \dfrac{3x}{3} < \dfrac{6}{3} \Rightarrow x < \dfrac{6}{3} \Rightarrow x < 2$$

Given the inequation $-2x < 4$, let's solve it: $$-2x < 4 \Rightarrow \dfrac{-2x}{-2} > \dfrac{4}{-2} \Rightarrow x > \dfrac{4}{-2} \Rightarrow x > -2$$

Now that we already know how to add and to subtract, as well as to multiply and to divide on both sides of an inequation by a concrete value, we are already able to solve any first degree inequation.