### Inequalities and linear programming - (8.1-8.3)

#### Solve compound linear inequalities in one unknown

• For any real number a and b

1) If $$a > b$$
Given $$x > a$$ and $$x > b$$
Then $$x > a$$
for value of x greater than b may not be greater than a,
the selected value range must satisfy both constraint connected by “and”)

2) If $$a > b$$
Given $$x > a$$ or $$x > b$$
Then $$x > b$$
for value of x either greater than a or greater than b
satisfies the constraint connected by “or” )
• Solve the following compound inequalities and represent the solutions graphically.
$$a)\qquad\left\{ \begin{array}{a} 4(x - 1) < x + 5\\6(2x + 3) < 11(x - 2)\end{array} \right.$$
$$b)\qquad x + 5 \le 2x + 7$$ or $$7x - 5 < 3x - 4$$
• $$a)\qquad 4(x - 1) < x + 5$$ and $$6(2x + 3) < 11(x - 2)$$
$$4x - 4 < x + 5$$ and $$12x + 18 < 11x - 22$$
$$3x < 9$$ and $$x < - 40$$
$$x < 3$$ and $$x < - 40$$
The solutions of the compound inequality are $$x < - 40$$.
Graphically,

$$b)\qquad x + 5 \le 2x + 7$$ or $$7x - 5 < 3x - 4$$
$$- x \le 2$$ or $$4x < 1$$
$$x \ge - 2$$ or $$x < \frac{1}{4}$$
The solutions of the compound inequality are all real numbers.
Graphically,

#### Solve quadratic inequalities in one unknown by the graphical method

• The figure shows the graph of$$\;y = \frac{4}{{25}}\left( {{x^2} - 14x + 24} \right)$$ .
Solve the quadratic equation $${x^2} - 14x + 24 < 0$$ graphically.

• From the graph $$y = \frac{4}{{25}}\left( {{x^2} - 14x + 12} \right),\;$$
For $$2 < x < 12$$ , $$y < 0$$
$$\frac{4}{{25}}\left( {{x^2} - 14x + 12} \right)\left\langle {0\; = } \right\rangle \;{x^2} - 14x + 12 < 0$$
$$\therefore$$ the solution to $${x^2} - 14x + 12 < 0$$ is $$2 < x < 12$$

#### Solve quadratic inequalities in one unknown by the algebraic method

• Solve the inequality $${x^2} - x - 56 \ge 0$$ by the algebraic method.
• $$\begin{array}{1}{x^2} - x - 56 \ge 0\\(x + 7)(x - 8) \ge 0\end{array}$$

$$\left\{ \begin{array}{l}x + 7 \ge 0\\x - 8 \ge 0\end{array} \right.$$ or $$\left\{ \begin{array}{l}x + 7 \le 0\\x - 8 \le 0\end{array} \right.$$

$$\left\{ \begin{array}{l}x \ge - 7\\x \ge 8\end{array} \right.$$ or $$\left\{ \begin{array}{l}x \le - 7\\x \le 8\end{array} \right.$$

$$x \ge 8$$ or $$x \le - 7$$
Therefore,the solutions of the inequality are $$x \le - 7$$ or $$x \ge 8$$.

Alternative method: $$\begin{array}{c}{x^2} - x - 56 \ge 0\\(x + 7)(x - 8) \ge 0\end{array}$$
Construct the following table to show the signs of the expressions of
$$x + 7$$, $$x - 8$$ and $$(x + 7)(x - 8)$$ in the relevant intervals of x.

$$x < - 7$$$$x = - 7$$$$- 7 < x < 8$$$$x = 8$$$$x > 8$$
$$x + 7$$-0+++
$$x - 8$$---0+
$$(x + 7)(x - 8)$$+0-0+

According to the table above, the solutions of the inequality are $$x \le - 7$$ or $$x \ge 8$$.