#### Chapter 3 Inequalities in one Variable

Section 3.2 Transformation of Inequalities

# 3.2.1 Transformation with Case Analysis

The simple linear transformations described in the previous section are equivalent transformations. They do not change the solution set of the corresponding inequality. For nonlinear inequalities, advanced solution methods are required. Usually, these methods need a case analysis depending on the sign. This is because, in contrast to the situation for equations described in Module 2, now also the comparator can be inverted when performing transformations.
##### Info 3.2.1

If an inequality is multiplied by a term in which the variable $x$ occurs, a case analysis is required and for each case the transformation has to be considered separately:
• For those values of $x$, for which the multiplied term is positive, the comparator of the inequality is unchanged.
• For those values of $x$, for which the multiplied term is negative, the comparator of the inequality is inverted.
• The case that the multiplied term is zero has to be excluded during the transformation and has to be considered separately, if necessary.

The solution sets found in the individual cases have to be checked with respect to the case conditions as described for the solution of absolute value equations.

In contrast, adding terms in which the variable occurs, does not require a case analysis. Usually, transformations involving case analyses are mandatory if the variable occurs in the denominator or in a composite term.
##### Example 3.2.2
The inequality $\frac{1}{2x}\le 1$ can be simplified by multiplying both sides of the inequality by the term $2x$:
• Under the condition $x>0$ this results in the new inequality $1\le 2x$. It has the solution set $L{}_{1}=\left[\frac{1}{2};\infty \left[\text{}$. The condition $x>0$ is satisfied by all elements of the solution set.
• Under the condition $x<0$ this results in the new inequality $1\ge 2x$. It has the solution set $\text{}\right]-\infty ;\frac{1}{2}\right]$. Because of the additional condition $x<0$ only the elements of the set $L{}_{2}=\text{}\right]-\infty ;0\left[\text{}$ are solutions.
• The single case $x=0$ is no solution since this value is not in the domain of the inequality. In this case multiplying the inequality by $x$ is not allowed.
So, altogether one obtains the union set $L=L{}_{1}\cup L{}_{2}=ℝ\setminus \left[0;\frac{1}{2}\left[\text{}$ as solution set: As in Module 2 the following statement holds for the solution set.
##### Info 3.2.3

The cases have to be chosen such that all elements of the domain of the inequality are covered. For the solution set in an individual case, it has to be checked that the solution set satisfies the corresponding case condition. For any case, the resulting solution set has to be reduced to the solution subset satisfying the case condition. The union of the solution sets for the individual cases is the solution set of the initial inequality.