1.2 Absolute value and inequalities

1 If $x - 2$ $<$ $5$, then adding $2$ on both sides leads to solution $x$ $<$ $7$.

2 Subtracting $3$ from both sides of the inequality $x$ $>$ $3$ $=$ $-6$, leads to solution $x$ $>$ $-9$.

3 Solving the inequality $1$ $\leq$ $2x$ $+$ $2$ $\leq$ $4$ means subtracting $2$ from all of the operands and then dividing by $2$. This leads to solution $-\frac{1}{2}$ $\leq$ $x$ $\leq$ $1$.

4 The inequality $3x$ $+$ $6$ $\geq$ $-3$ can be solved by first dividing by $3$ and then subtracting $2$ from the inequality. This gives the solution $-3$ $\geq$ $x$.

5 Given the inequality $3$ $<$ $2 - 5x$ $<$ $13$, one can subtract $-2$ to get $1$ $<$ $-5x$ $<$ $11$ and then divide by $-5$ to get the solution $-\frac{1}{5}$ $>$ $x$ $>$ $-\frac{11}{5}$.

6 When every element of inequality $-4$ $<$ $\frac{2x-4}{3}$ $\leq$ $7$ is multiplied by $3$, added with $4$ and divided by $2$, the inequality can be written as $-4$ $<$ $x$ $\leq$ $\frac{11}{2}$, which is the solution (set) for the given inequality.

7 To get the solution for inequality $1$ $-$ $2x$ $<$ $-2$, first the inequality can be multiplied by $-1$ to get the signs flipped in the inequality, so that it becomes $-1 + 2x$ $>$ $2$. Then $1$ can be added to both sides and finally both of the terms can divided by $3$ to get solution the $\frac{2}{3}$ $<$ $x$.

8 The inequality $0$ $<$ $\frac{5 - 4x}{2}$ $<$ $1$, can multiplied with $2$, subtracted by $-5$, multiplied with $-1$ and the divided by $-4$ to get the solution $\frac{5}{4}$ $>$ $x$ $>$ $\frac{3}{4}$.

9 Adding $5$ to and dividing by $3$ the terms in the inequality $1$ $\leq$ $3x$ $-$ $5$ $<$ $3$ reveals the solution $\frac{5}{3}$ $\leq$ $x$ $<$ $\frac{8}{3}$.

10 If a inequality is $2$ $\geq$ $\frac{4-2x}{5}$ $>$ $-4$, then it can be solved by first multiplying the inequalities by $5$ and subtracting $-4$ from all elements to get inequality $10$ $\geq$ $-2x$ $>$ $-20$. Then all of the elements can be multiplied by $-2$ (note that this flips the inequalities) to get the solution $-5$ $\leq$ $x$ $<$ $10$.

11 Multiplying terms of the inequality $\frac{2}{5x}$ $<$ $4$, by $5x$ leads to inequality $2$ $<$ $4(5x)$ $=$ $20x$. Dividing both sides with $20$ leads to solution $\frac{1}{10}$ $<$ $x$.

12 …

13 Multiplying both sides of the inequality $\frac{2}{5x}$ $<$ $4$ by $5x$ gives $2$ $<$ $4(5x)$, which can be divided by $4$ to get $\frac{1}{2}$ $<$ $5x$. Then dividing by $5$, it can be seen that $\frac{1}{10}$ $<$ $x$.

14 Because $x^2$ $-$ $2x$ $-$ $8$ $=$ $(x+2)(x-4)$, the critical points are $x$ $=$ $-2$ and $x$ $=$ $4$, so there are three intervals of real values to test: $x$ $<$ $-2$, $-2$ $\leq$ $x$ $\leq$ $4$ and $x$ $>$ $4$.

14 If $x^2$ $-$ $2x$ $-$ $8$ $=$ $0$ is equivalent to $(x+2)$$(x-4)$ $=$ $0$, so the zero-points are at $x$ $=$ $-2$ and $x$ $=$ $4$. This leads to three intervals to test $x$ $<$ $-2$, $-2$ $\leq$ $x$ $\leq$ $4$ and $x$ $>$ $4$. When $x$ is a number between $-2$ and $4$, such as $x$ $=$ $0$, only the inequality $-2$ $\leq$ $x$ $\leq$ $4$ is true, so the solutions to the inequality lies on that interval.

15 Because $x^2$ $-$ $5x$ $\geq$ $14$ is equal to $x^2$ $-$ $5x$ $-$ $14$ $\leq$ $0$, the inequality can be solved in similar manner as it was done in 14, that is first by finding the roots when $x^2$ $-$ $5x$ $-$ $14$ $=$ $0$. Because $x^2$ $-$ $5x$ $-$ $14$ $=$ $(x+7)$$(x-2)$, the solutions for when the equation is $0$, are when $x$ is $-7$ or $2$. Testing the three possible intervals $x$ $<$ $-7$, $-7$ $\leq$ $x$ $\leq$ $0$ and $x$ $>$ $2$. Picking a number between the roots, e.g. when $x$ $=$ $0$, shows that the only the second inequality is true, so the solutions for the inequality must lie there, that is, at the interval $-7$ $\leq$ $x$ $\leq$ $2$.

16 Solving the inequality $a$ $\leq$ $\frac{bx + c}{d}$ $<$ $e$ can be done by first multiplying all terms with $d$, then dividing the terms by $b$ and finally subtracting $-c$. This leads to solution $\frac{ad}{b}$ $-$ $c$ $\leq$ $x$ $<$ $\frac{de}{b}$ $-$ $c$ equivalent to $a\frac{d}{b}$ $-$ $c$ $\leq$ $x$ $<$ $e\frac{d}{b}$ $-$ $c$.

17 (a) Solving for $x$: $x$ $=$ $|4 - 5|$ $=$ $|-1|$ $=$ $|1|$ $=$ $1$, (b) similarly $x$ $=$ $|-6 - (-2)|$ $=$ $|-6 + 2|$ $=$ $|-4|$ $=$ $|4|$ $=$ $4$ and (c) $x$ $=$ $|2|$ $-$ $|-3|$ $=$ $|2|$ $-$ $|3|$ $=$ $2$ $-$ $3$ $=$ $1$.

18 …

19 If $x$ and $y$ are nonnegative, then $|x + y|$ $=$ $(x + y)$ by how the absolute value (function) is defined and $|x|$ $+$ $|y|$ $=$ $x + y$, so taken together $|x + y|$ $=$ $(x + y)$ $=$ $x + y$ $=$ $|x|$ $+$ $|y|$, that is, $|x + y|$ $=$ $|x|$ $+$ $|y|$.

20 If $x$ $>$ $0$ and $y$ $<$ $0$, then $|x|$ $=$ $x$ and $|y|$ $=$ $|-y|$ $=$ $-(-y)$ $=$ $y$ by the definition of the absolute value, then $|x|$ $+$ $|y|$ $=$ $x$ $+$ $y$. Moreover, because $|x+y|$ $=$ $|x+(-y)|$ $=$ $|x-y|$, that is either equal to $x-y$ or $-(x-y)$ $=$ $-x+y$, which are both strictly less than $|x|$ $+$ $|y|$ $=$ $x+y$, then $|x+y|$ $<$ $|x| + |y|$.

21 If $x$ $<$ $0$ and $y$ $<$ $0$, then $|x|$ $=$ $|-x|$ $=$ $-(-x)$ $=$ $x$ and $|y|$ $=$ $|-y|$ $=$ $-(-y)$ $=$ $y$, so $|x|$ $+$ $|y|$ $=$ $x+y$, and $|x+y|$ $=$ $|(-x) + (-y)|$ $=$ $|-(x+y)|$ $=$ $-[-(x+y)]$ $=$ $x+y$, then $|x|$ $+$ $|y|$ $=$ $x+y$ $=$ $-[-(x+y)]$ $=$ $|-(x+y)|$ $=$ $|(-x)$ $+$ $(-y)|$ $=$ $|x+y|$. Because of this, $|x+y|$ $=$ $|x|$ $+$ $|y|$, when $x$ $<$ $0$ and $y$ $<$ $0$.

22 Using the proofs of 19 to 21, it can be seen that for any real number $x$ and $y$, $|x+y|$ $\leq$ $|x|$ $+$ $|y|$.

23 By the triangle inequality, $|x-y|$ $\leq$ $|x|$ $-$ $|y|$ and $|y-x|$ $\leq$ $|y|$ $-$ $|x|$ $=$ $-|x|$ $+$ $|y|$, so

$$ -|x| + |y| \leq |x-y| \leq |x| + |y|, $$

which by the definition of absolute value is equal to $\big||x|-|y|\big|$ $\leq$ $|x-y|$.