Inequalities and absolute value

Inequalities work like equations, with one exception

You may add or subtract any number on both sides of an inequality, and multiply or divide by any positive number, and the direction stays the same. Multiplying or dividing by a negative number reverses it:

$$-3x + 7 > 19 \;\Rightarrow\; -3x > 12 \;\Rightarrow\; x < -4$$

Check the result with one number: $x = -5$ gives $15 + 7 = 22 > 19$, as it should.

The exception has a consequence that catches many people: you cannot multiply or divide by a variable whose sign you do not know. From $\frac{1}{x} < 2$ you cannot conclude $1 < 2x$, because if $x$ is negative the direction flips. Instead, split into cases ($x > 0$ and $x < 0$) or test numbers.

Two other operations need the same care. Squaring both sides is safe only when both sides are non-negative ($-3 < 2$, but $9 > 4$). Taking reciprocals of two numbers with the same sign reverses the direction ($2 < 5$, but $\frac{1}{2} > \frac{1}{5}$).

Compound inequalities

A statement such as $-1 < 2x + 3 \le 11$ is two inequalities at once. Do the same thing to all three parts: subtract 3 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 ($\le$); that matters the moment the question counts integers.

Absolute value is distance

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

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

So $|x - 2| > 5$ means $x$ lies more than 5 away from 2: $x < -3$ or $x > 7$. Notice the word or: two separate rays, never one interval. A "less than" statement gives one interval in the middle; a "greater than" statement gives the two outside pieces.

For $|x + 3|$, rewrite it as $|x - (-3)|$: the centre is $-3$, not 3.

Absolute value equations: cases, then check

When the expression inside is not simply $x - a$, solve by cases. For $|E| = k$, write $E = k$ or $E = -k$ and solve each. If the right-hand side contains the variable, as in $|x + 3| = 2x$, a case can produce a value that does not satisfy the original equation, since an absolute value can never equal a negative number. Substitute every candidate back and discard the false ones.

Counting integers in an interval

From $-\frac{7}{3} < x < 5$, the integers are $-2, -1, 0, 1, 2, 3, 4$: seven of them. Two habits prevent off-by-one errors. Convert awkward endpoints to decimals ($-\frac{7}{3} \approx -2.33$, so $-2$ is the first integer above it). And check each end separately for whether it is included: a strict $< 5$ excludes 5.

Ranges of combinations

If you know a range for $a$ and a range for $b$, you can bound combinations:

• Sum: add the lower ends and add the upper ends.
• Difference $a - b$: the largest value is the top of $a$ minus the bottom of $b$; the smallest is the bottom of $a$ minus the top of $b$.
• Product: check all four products of the endpoints. Because of signs, the largest product can come from two negatives rather than two positives.

These questions reward a quick table of corner values rather than intuition.