Chapter 5 - Linear Inequalities

Multiplying an inequality by a negative number reverses the sign.
Given $x < 5$, multiplying by $-1$ gives $-x > -5$.

Multiplying an inequality by a negative number reverses the sign.
Given $x \le 5$, multiplying by $-1$ gives $-x \ge -5$.

$-3x + 17 < -13 \implies -3x < -13 - 17 \implies -3x < -30$.
Dividing by $-3$ reverses the inequality: $x > 10$.
So $x \in (10, \infty)$.

The inequality $|x| < a$ implies $-a < x < a$.
Here $|x| < 3 \implies -3 < x < 3$.

The inequality $|x| > a$ implies $x < -a$ or $x > a$.
Here $|x| > b \implies x < -b$ or $x > b$.
In interval notation: $x \in (-\infty, -b) \cup (b, \infty)$.

$|x - 1| > 5 \implies x - 1 < -5$ or $x - 1 > 5$.
$x < -4$ or $x > 6$.
So $x \in (-\infty, -4) \cup (6, \infty)$.

$|x + 2| \le 9 \implies -9 \le x + 2 \le 9$.
Subtracting 2 from all parts: $-11 \le x \le 7$.
So $x \in [-11, 7]$.

The graph shows the region outside $-5$ and $5$ with open circles, meaning $x < -5$ or $x > 5$.
This corresponds to $|x| > 5$.

The graph shows values greater than 5 with an open circle at 5.
This represents the interval $(5, \infty)$.

The graph shows values less than or equal to 5 (closed circle).
This represents the interval $(-\infty, 5]$.

The graph shows values greater than or equal to 5 (closed circle).
This represents the interval $[5, \infty)$.

The graph shows values less than 5 with an open circle.
This represents the interval $(-\infty, 5)$.

(i) False. Dividing by a negative number reverses the inequality. $\frac{x}{b} > \frac{y}{b}$.
(ii) False. If $xy > 0$, both $x, y$ are positive OR both are negative.
(iii) False. If $xy < 0$, they have opposite signs.
(iv) True. Intersection of $x > 5$ and $x > 2$ is $x > 5$.
(v) True. $|x| < 5 \implies -5 < x < 5$.
(vi) False. Strict inequality ($>$) does not include the boundary line.
(vii) False. Substitute $(1, 0)$: $1 - 0 = 1 \not\le 0$.

(i) Divide by $-4$ (negative), reverse sign: $x \le -3$.
(ii) Multiply by $-4/3$ (negative), reverse sign: $x \ge (-3)(-4/3) = 4$.
(iii) Since numerator $2 > 0$, denominator must be positive: $x+2 > 0 \implies x > -2$.
(iv) Multiply by 4 (positive), sign stays: $4x > -20$.
(v) $x > y, z < 0 \implies xz < yz$ (multiply by negative). Then multiply by $-1$: $-xz > -yz$.