### Solving equations and inequalities: Linear inequalities in one unknown

### Solving a system of inequalities with one unknown

In the margin we have already noted that for example \(-1<x<1\) is a compact notation for \(-1<x\) **and** \(x<1\), also denoted with the logical operator \(\wedge\) for 'and' as \(-1<x\;\wedge\;x<1\). Because \(-1<x\) is equivalent to \(x>-1\) the statement \(-1<x<1\) means '\(x\) greater than \(-1\) and less than \(1\)'. A more comprehensive notation for this is a system of inequalities; in this example: \[\left\{\;\begin{aligned} x &> -1 \\ x&<1\end{aligned} \right.\]

In general, the solution of a **system of inequalities** **with one unknown** is a description of all values of this unknown that make all inequalities true statements after substitution. The system \[\lineqs{x+2 &\ge& 3\cr 4x-5&<&6\cr}\] is, for example, a different notation for \[(x+2\ge3)\;\wedge\;(4x-5<6)\] The **solution** consists of a description of the conditions which \(x\) must satisfy in a form in which the variable \(x\) is isolated. In this example: \[(x\ge1)\;\wedge\;(x<\frac{11}{4})\] and we often write it shorter as \[1\le x<\frac{11}{4}\] Another notation for the solution set, in which the unknown is not present anymore, is that of a semi-open interval \([1,\tfrac{11}{4})\) or \([1,2\tfrac{3}{4})\).

Solving by reduction When solving a system of inequalities with one unknown by reduction

- you first solve each inequality separately by reduction,
- next, you combine the intermediate results with the logical operator \(\wedge\) for 'and',
- and finally you possibly simplify the logical expression.

Solving via equations When solving a system of inequalities with one unknown via equations

- you first solve each inequality separately via equations,
- next, you combine
- the intermediate results with the logical operator \(\wedge\) for 'and'
- and finally you possibly simplify the logical expression.

After all,

when you solve the two inequalities separately, you get \[\lineqs{x&\gt& {{5}\over{4}}\cr x&\ge&-{{3}\over{2}}\cr}\] Because \(-{{3}\over{2}}<{{5}\over{4}}\), it follows that the solution consists only of values of \(x\) greater than \({{5}\over{4}}\).