Linear Inequations

Samia CHALLAL

19 Linear Inequations

$\bullet$ A linear inequality is a statement in the form $a x + b < 0$ , or $a x + b > 0$ , or $a x + b \leqslant 0$ , or $a x + b \geqslant 0$ , where $x$ is a variable.

The variable that makes the statement true is called a solution of the equation.

$\bullet$ How to solve, for example, $a x + b < 0$ ?

$-$ combine all variable terms in one side

$-$ combine all constants terms on the other

$-$ divide both sides by the coefficient of the
variable and change the inequality to $">"$ if $a<0$ .

$\bullet$
$a x + b < 0\quad\Longleftrightarrow\quad a x +b + (-b) < 0+(- b) \quad\Longleftrightarrow\quad a x < - b$

$\quad \Longleftrightarrow \quad \left\{ \begin{matrix} x< - \frac{b}{a}\quad & \hbox { if } \quad a> 0 \\ \\ x > - \frac{b}{a}\quad & \hbox { if } \quad a< 0 \\ \\ \hbox{ no solution } \quad & \hbox { if } \quad a = 0 \quad \hbox { and } \quad b> 0\\ \\ \hbox{ infinite solution } \quad & \hbox { if } \quad a =0 \quad \hbox { and } \quad b< 0 \end{matrix} \right.$

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Exercise 1

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The set of possible solutions is the interval: $\displaystyle{ I= [-\frac{37}{3}, +\infty)$ .

Exercise 2

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The set of possible solutions is the interval: $\displaystyle{ I= \Big(\frac{1}{3}, \frac{49}{9}\Big]$ .

Exercise 3

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We have

$\displaystyle{- \Big(\frac{y+8}{2} \Big) + 3(y + 1 )\leq -1 + \frac{5}{2} y }$

$\Longleftrightarrow\qquad \displaystyle{ \Big(-\frac{1}{2} + 3 -\frac{5}{2} \Big) y \leq -1 -3 + 4 }$

$\Longleftrightarrow\qquad \displaystyle{ \Big(0 \Big) y \leq 0 }$

which is true for any $y\in \mathbb{R}$ since $\Big(0 \Big) y =0 \leq 0$ $\forall y$ .

The set of all possible solutions is the interval: $\displaystyle{ I= (-\infty, +\infty)$ .

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