Inequalities and absolute value

Inequalities behave like equations, with one exception

You may add or subtract the same number on both sides of an inequality, and you may multiply or divide both sides by the same positive number, without changing the direction of the sign. The exception: multiplying or dividing by a negative number reverses the sign.

Why? Multiplying by $-1$ reflects every number across zero on the number line, so the order of any two numbers flips. $2 < 5$, but $-2 > -5$.

$$7 - 3x \ge 19 \;\Rightarrow\; -3x \ge 12 \;\Rightarrow\; x \le -4$$

The sign flipped at the division by $-3$, and only there. Subtracting 7 did not flip it. To check the direction, test one value on each side of the boundary in the original inequality: $x = -5$ gives $22 \ge 19$ (true) and $x = 0$ gives $7 \ge 19$ (false), so the solution really is $x \le -4$.

Never multiply by a variable of unknown sign

From $\frac{6}{x} > 2$ you cannot simply write $6 > 2x$, because if $x$ is negative the sign should flip. Unless the stem tells you the sign of $x$, treat the cases separately or test values. Here, a negative $x$ makes $\frac{6}{x}$ negative, which is never greater than 2, so $x$ must be positive, and then $6 > 2x$ gives $0 < x < 3$.

Compound inequalities

An inequality such as $-3 < 2x + 1 \le 9$ is two inequalities at once. Do the same thing to all three parts: subtract 1 to get $-4 < 2x \le 8$, then divide by 2 to get $-2 < x \le 4$. Keep track of which ends are strict ($<$) and which include the endpoint ($\le$); a question that asks for the number of integers in a range turns on this.

Combining ranges

If $-3 < x < 2$ and $1 < y < 5$, what values can $x + y$ and $x - y$ take?

• Sum: add the lower ends and add the upper ends: $-2 < x + y < 7$.
• Difference: the smallest $x - y$ uses the smallest $x$ and the largest $y$; the largest uses the largest $x$ and the smallest $y$. So $-3 - 5 < x - y < 2 - 1$, that is $-8 < x - y < 1$.

Subtracting the lower ends from each other ($-3 - 1 = -4$) and the upper ends from each other ($2 - 5 = -3$) gives a range that is too narrow and is a frequent wrong answer. Products are harder still, because signs interact; with mixed signs, test the four corner products.

Absolute value is distance

$|x|$ is the distance from $x$ to $0$ on the number line, so it is never negative. More usefully, $|x - a|$ is the distance from $x$ to $a$. That reading turns absolute value inequalities into pictures:

Statement	Meaning	Solution
$\lvert x - a \rvert < b$	within $b$ of $a$	$a - b < x < a + b$
$\lvert x - a \rvert > b$	more than $b$ from $a$	$x < a - b$ or $x > a + b$
$\lvert x - a \rvert = b$	exactly $b$ from $a$	$x = a - b$ or $x = a + b$

For $|x + 4| \le 3$, rewrite $x + 4$ as $x - (-4)$: the points within 3 of $-4$, so $-7 \le x \le -1$. If the inside has a coefficient, as in $|2x - 5| \le 9$, write the compound form $-9 \le 2x - 5 \le 9$ and solve all three parts.

Absolute value equations: two cases and a check

$|E| = k$ means $E = k$ or $E = -k$. When $k$ is a plain number, both cases are genuine solutions as long as $k \ge 0$; if $k$ is negative, there is no solution at all. When the right side contains the variable, as in $|x + 1| = 2x - 4$, solve both cases and then substitute each answer into the original: a case can produce a value that makes the right side negative, and an absolute value can never equal a negative number. Such a value is extraneous and must be thrown out.