#### Chapter 3 Inequalities in one Variable

Section 3.1 Inequalities and their Solution Sets

# 3.1.2 Solving simple Inequalities

If the variable occurs isolated in the inequality, the solution set is an interval, see also info box 1.1.5:
##### Info 3.1.3

The solved inequalities have the following intervals as their solution sets:
• $x has the solution set $\text{}\right]-\infty ;a\left[\text{}$, i.e. all $x$ less than $a$.
• $x\le a$ has the solution set $\text{}\right]-\infty ;a\right]$, i.e. all $x$ less than or equal to $a$.
• $x>a$ has the solution set $\text{}\right]a;\infty \left[\text{}$, i.e. all $x$ greater than $a$.
• $x\ge a$ has the solution set $\left[a;\infty \left[\text{}$, i.e. all $x$ greater than or equal to $a$.
Here, $x$ is the variable and $a$ is a specific value. If the variable does not occur in the inequality anymore, the solution set is either $ℝ=\text{}\right]-\infty ;\infty \left[\text{}$ if the inequality is satisfied, or the empty set $\left\{\right\}$ if the inequality is not satisfied.

The symbol $\infty$ means infinity. A finite interval has the form $\text{}\right]a;b\left[\text{}$ which reads "all numbers between $a$ and $b$". If the interval is bounded only on one side, we can write the symbol $\infty$ (right-hand side) or $-\infty$ (left-hand side) as the other bound.
As in the case of equations, one tries to find a solved inequality by applying transformations that do not change the solution set. The solution set can be read off from the solved inequality.
##### Info 3.1.4

To obtain a solved inequality from an unsolved inequality the following equivalent transformations are allowed:
• adding a constant to both sides of the inequality: $a is equivalent to $a+c.
• multiplying both sides of the inequality by a positive constant: $a is equivalent to $a·c if $c>0$.
• multiplying both sides of the inequality by a negative constant and inverting the comparator: $a is equivalent to $a·c>b·c$ if $c<0$.

##### Example 3.1.5
The inequality $-\frac{3}{4}x-\frac{1}{2}<2$ is solved stepwise by the above transformations:

So, the initial inequality has the solution set $\text{}\right]-\frac{10}{3};\infty \left[\text{}$. Importantly, multiplying the inequality by the negative number $-\frac{4}{3}$ inverts the comparator.

##### Exercise 3.1.6
Are the following inequalities true or false?
 $\frac{1}{2}>1-\frac{1}{3}$ ${a}^{2}\ge 2ab-{b}^{2}$ (where $a$ and $b$ are unknown numbers) $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}$ Let $a, then also ${a}^{2}<{b}^{2}$.

##### Exercise 3.1.7
Find the solution sets of the following inequalities.
1. $2x+1>3x-1$ has the solution interval $L$$=$ .
2. $-3x-\frac{1}{2}\le x+\frac{1}{2}$ has the solution interval $L$$=$ .
3. $x-\frac{1}{2}\le x+\frac{1}{2}$ has the solution interval $L$$=$ .
Enter the intervals in the form $\left[a;b\right]$, $\text{}\right]a;b\right]$, etc., for the interval boundaries also fractions and infinity or -infinity can be used. Take care whether the interval boundaries are included or excluded.

##### Info 3.1.8

An inequality in one variable $x$ is linear if on both sides of the inequality only multiples of $x$ and constants occur. Each linear inequality can be transformed into a solved inequality by one of the equivalent transformations described in the info box 3.1.3.