Absolute Value & Inequalities

What you need to know

• Absolute value is distance. $|x-a|$ is how far $x$ is from $a$ on the number line. Example: $|7-3|=|-1-3|=4$, since $7$ and $-1$ are both $4$ away from $3$.
• Three cases (for a positive number $c$). $|x|=c$ means $x=c$ or $x=-c$. $|x|<c$ means $-c<x<c$. $|x|>c$ means $x>c$ or $x<-c$. Example: $|x|>3$ holds for $5$ and $-4$, but not for $2$.
• Several absolute values. A breakpoint is an $x$ that makes one of the insides $0$. Between breakpoints each inside keeps one sign, so each absolute value can be dropped (with a $+$ or $-$ in front). The graph is straight pieces that bend only at breakpoints. Example: $|x-1|+|x-3|$ is $4-2x$ for $x\le 1$, $2$ for $1\le x\le 3$, and $2x-4$ for $x\ge 3$.
• Largest and smallest values. If such an expression has a smallest or largest value, it happens at a breakpoint. If $x$ is limited to an interval, also check the ends. Example: $|x-1|+|x-3|$ is smallest, $2$, at the breakpoints $1$ and $3$ (and between them).
• AM-GM. For $a$ and $b$ zero or positive, the average is at least the geometric mean: $\dfrac{a+b}2\ge\sqrt{ab}$, with equality only when $a=b$. So a sum with a fixed product is smallest, and a product with a fixed sum is largest, when the terms are equal. Example: for $x>0$, $x+\frac4x\ge 2\sqrt4=4$, reached at $x=2$.
• Inequality rules. Multiplying both sides by a negative number reverses the inequality: $-2x<6$ becomes $x>-3$. Squaring is safe only when both sides are zero or positive: $-3<2$, but $9>4$.

How AMC 10 tests it

• Problems 4–10: solve $|2x-5|=7$ or $|x-1|+|x-3|=4$ and give the sum or the number of solutions.
• Problems 8–14: the region $|x|+|y|\le c$ is a diamond; find its area or count its lattice points (points with integer coordinates).
• Problems 10–16: find the smallest value of $|x-1|+|x-2|+\dots+|x-n|$ (it is at the median) or of $x+\frac4x$ (AM-GM).
• Problems 12–18: count the integers $n$ that satisfy a chain such as $\sqrt n<\frac k3<\sqrt{n+1}$, by bounding carefully.

Standard approaches

1. Read one or two absolute values as distances on a number line; often you can see the answer. Example: $|x-1|+|x-3|=4$ at $x=0$ and $x=4$.
2. Otherwise, split into cases at the breakpoints. Solve each piece, and check that each solution lies in its own interval.
3. When $|x|$ and $|y|$ both appear, graph the first quadrant, then reflect across both axes.
4. For a largest or smallest value, try AM-GM first (check that equality can happen), then complete the square.
5. To count integers, get the variable alone and find the exact endpoints. Let the wording ("less than" or "at most") decide if each endpoint counts.

Worked example

What is the area of the region of the plane consisting of all points $(x,y)$ with $|x-y|+|x+y|\le4$?

(A) $8$ (B) $8\sqrt2$ (C) $16$ (D) $16\sqrt2$ (E) $32$

Start with the wedge on the right where $x\ge |y|$. There $x-y$ and $x+y$ are zero or positive, so we can drop the absolute values:
$$(x-y)+(x+y)\le4,\quad\text{so}\quad x\le2.$$
This piece is the triangle with corners $(0,0)$, $(2,2)$ and $(2,-2)$. Its area is $\frac12\cdot4\cdot2=4$ (base $4$, height $2$).

By symmetry, the other three wedges ($x\le -|y|$, $y\ge |x|$, $y\le -|x|$) give three matching triangles. The total area is $4\cdot4=16$.

Another way: $|x-y|+|x+y|$ always equals twice the larger of $|x|$ and $|y|$. So the region is the square where $|x|\le 2$ and $|y|\le 2$, with area $4^2=16$.

The answer is $\boxed{\textbf{(C)}\ 16}$.

Pitfalls

• Keeping a case solution that lies outside its case. Example: in $|x-1|+|x-3|=1$, the case $x\ge 3$ gives $x=2.5$, which is not $\ge 3$; throw it out.
• Solving $|x|>c$ as $-c<x<c$. That mixes up "or" with "and".
• Using AM-GM on numbers that may be negative ($x=-1$ gives $x+\frac4x=-5$, less than $4$), or not checking that equality can happen.
• Multiplying an inequality by a variable whose sign you don't know, without splitting into cases. Example: $\frac1x<1$ does not mean $x>1$, because $x=-2$ works too.

Traps that recur

• Reporting the area of one quadrant's diamond (2) without multiplying by 4 for the four quadrants. (2023 AMC 10B #13)
• Assuming the parabola crosses the inner diamond's edges in extra points, or miscounting the tangency-type contacts at the vertices (0,3), (3,0), (-3,0). (2021 AMC Fall 10A #14)
• Forgetting the 100 integers where P(n) = 0 and answering 5000, or choosing the gaps with even k instead of odd k. (2020 AMC 10A #17)
• Removing the outer absolute value as |x| - |y| = 1 only, missing the branch |y| - |x| = 1 that produces the third solution (0, 1). (2018 AMC 10A #12)