Solving a system of inequalities with one unknown

Solving linear 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 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.

Determine the exact solution of the following system of inequalities $\lineqs{10x-6 &\ge& 0\cr 3x+3&\ge &0\cr}$

$x \ge {{3}\over{5}}$

After all,
when you solve the two inequalities separately, you get $\lineqs{x&\ge& {{3}\over{5}}\cr x&\ge&-1\cr}$ Because $-1<{{3}\over{5}}$ , it follows that the solution consists only of values of $x$ greater than or equal to ${{3}\over{5}}$ .

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