Absolute Value Equations and Graphs, SAT math lesson

Stand at $3$ on a number line and walk $5$ units. You end at $8$ or at $-2$ , depending on the direction. An absolute value equation asks exactly this kind of question, and that is why it usually has two answers. Seeing the distance behind the symbols tells you when it has two, one or none, and how to keep from accepting a wrong one.

See also Absolute Value as Distance and a Graph in Foundations for Calculus and Absolute Value Functions, Equations, and Inequalities in Precalculus.

Key terms

Absolute value

Definition The absolute value of a real number $a$ is

\lvert a \rvert = \begin{cases} a, & a \ge 0, \\ -a, & a < 0. \end{cases}

For every real $a$ it satisfies $\lvert a \rvert \ge 0$ , with $\lvert a \rvert = 0$ only when $a = 0$ . It also satisfies $\lvert -a \rvert = \lvert a \rvert$ .

In plain words Absolute value keeps the size of a number and drops its sign. It can never be negative, so an equation that sets an absolute value equal to a negative number has no solution.

Example The values are $\lvert 7 \rvert = 7$ and $\lvert -7 \rvert = -(-7) = 7$ . For an expression, evaluate inside first. At $x = 1$ the value of $\lvert 2x - 9 \rvert$ is $\lvert -7 \rvert = 7$ , not $2 \cdot 1 + 9$ .

Distance

Definition For real numbers $a$ and $b$ , the distance between $a$ and $b$ on the number line is $\lvert a - b \rvert$ . In particular, $\lvert x \rvert$ is the distance from $x$ to $0$ , and for $d > 0$

\begin{gathered} \lvert x - c \rvert = d \quad \text{means} \\[0.4em] x = c + d \ \text{or} \ x = c - d. \end{gathered}

The same picture explains inequalities. For $d > 0$ , $\lvert x - c \rvert < d$ means $c - d < x < c + d$ , and $\lvert x - c \rvert > d$ means $x < c - d$ or $x > c + d$ .

In plain words Read $\lvert x - c \rvert$ as “how far $x$ is from $c$ .” Exactly two points sit at a given positive distance from $c$ , one on each side.

Example Solve $\lvert x - 4 \rvert = 3$ . The numbers at distance $3$ from $4$ are $4 + 3 = 7$ and $4 - 3 = 1$ . Both check, since $\lvert 7 - 4 \rvert = 3$ and $\lvert 1 - 4 \rvert = 3$ . The inequality $\lvert x - 4 \rvert \le 3$ describes everything between them, $1 \le x \le 7$ .

Case split

Definition Let $E$ be an expression in $x$ and let $d$ be a constant.

If $d > 0$ , then $\lvert E \rvert = d$ holds exactly when $E = d$ or $E = -d$ .
If $d = 0$ , then $\lvert E \rvert = 0$ holds exactly when $E = 0$ .
If $d < 0$ , then $\lvert E \rvert = d$ has no solution.

Before splitting, isolate the absolute value on one side so that the equation has the form $\lvert E \rvert = d$ .

In plain words The inside is either the positive distance or its negative. Solve both linear equations. A zero distance gives one case and a negative distance gives none.

Example Solve $\lvert 2x - 3 \rvert = 5$ . The two cases are

\begin{aligned} 2x - 3 &= 5 \quad \text{so} \quad x = 4, \\[0.4em] 2x - 3 &= -5 \quad \text{so} \quad x = -1. \end{aligned}

Check them. The values are $\lvert 8 - 3 \rvert = 5$ and $\lvert -2 - 3 \rvert = 5$ , so the solutions are $x = 4$ and $x = -1$ . Now compare $\lvert x - 1 \rvert + 6 = 2$ . Isolating gives $\lvert x - 1 \rvert = -4$ , and no distance is negative, so there is no solution. Splitting blindly would give $x = -3$ and $x = 5$ , and neither one works.

Piecewise interpretation

Definition Any expression $\lvert mx + n \rvert$ with $m \ne 0$ can be written without absolute value bars by splitting at the input where the inside is zero, $x = -\frac{n}{m}$ . On the side where $mx + n \ge 0$ it equals $mx + n$ , and on the side where $mx + n < 0$ it equals $-(mx + n)$ .

In plain words An absolute value function is two linear pieces glued together at the point where the inside changes sign.

Example The inside of $\lvert 2x - 3 \rvert$ is zero at $x = \frac{3}{2}$ , so

\lvert 2x - 3 \rvert = \begin{cases} 2x - 3, & x \ge \frac{3}{2}, \\[0.3em] -2x + 3, & x < \frac{3}{2}. \end{cases}

At $x = 4$ the first piece gives $5$ , and at $x = -1$ the second piece gives $2 + 3 = 5$ , matching the case split above.

Vertex

Definition For real numbers $a \ne 0$ , $h$ and $k$ , the graph of

y = a \lvert x - h \rvert + k

is a V shape with vertex $(h, k)$ . It opens upward when $a > 0$ and downward when $a < 0$ , and its two straight sides have slopes $a$ and $-a$ . The graph is symmetric about the vertical line $x = h$ . When $a > 0$ , the horizontal line $y = c$ meets the graph in two points if $c > k$ , in one point (the vertex) if $c = k$ , and in no point if $c < k$ .

In plain words The vertex is the corner of the V, where the inside of the absolute value is zero. Counting where a horizontal line crosses the V counts the solutions of the matching equation.

Example The graph of $y = \lvert x + 2 \rvert - 4$ has vertex $(-2, -4)$ and opens upward. Its $x$ -intercepts solve $\lvert x + 2 \rvert = 4$ , so $x = 2$ or $x = -6$ , which sit $4$ units on each side of $x = -2$ . In the Desmos graphing calculator you can type y=abs(x+2)-4, then y=0, and click the two intersection points to read $(-6, 0)$ and $(2, 0)$ . For the earlier equation, graphing y=abs(2x-3) and y=5 shows a V with vertex $\left(\frac{3}{2}, 0\right)$ crossing the line at $(-1, 5)$ and $(4, 5)$ .

Branch condition

Definition When the other side depends on $x$ , as in $\lvert E \rvert = g(x)$ , the equation can hold only where $g(x) \ge 0$ . It is equivalent to

\begin{gathered} E = g(x) \ \text{or} \ E = -g(x), \\[0.4em] \text{together with} \ g(x) \ge 0. \end{gathered}

Equivalently, use the piecewise form. On the branch where $E \ge 0$ solve $E = g(x)$ , on the branch where $E < 0$ solve $-E = g(x)$ , and keep a solution only if it lies on its own branch. Either way, check every candidate in the original equation.

In plain words With a variable on the right, the two cases are not automatically both valid. A candidate that makes the right side negative, or that falls outside its branch, must be thrown out.

Example Solve $\lvert x - 3 \rvert = 2x$ . On the branch $x \ge 3$ the equation is $x - 3 = 2x$ , so $x = -3$ , which is not on this branch. On the branch $x < 3$ it is $3 - x = 2x$ , so $x = 1$ , which is on this branch. Check in the original equation.

Candidate	$\lvert x - 3 \rvert$	$2x$	Valid
$x = -3$	$6$	$-6$	no
$x = 1$	$2$	$2$	yes

The only solution is $x = 1$ .

Check yourself

What is the sum of all solutions to $\lvert 3x - 4 \rvert = 11$ ?

Show the answer

The answer is $\frac{8}{3}$ . The case $3x - 4 = 11$ gives $x = 5$ , and the case $3x - 4 = -11$ gives $x = -\frac{7}{3}$ . Both check, and their sum is

5 - \frac{7}{3} = \frac{8}{3}.

Entering $5$ forgets the negative case. You could enter 8/3 or a decimal such as 2.666 or 2.667.

For which value of $c$ does the equation $\lvert x + 3 \rvert - 2 = c$ have no real solution? (A) $-3$ , (B) $-2$ , (C) $0$ , (D) $5$

Show the answer

The correct choice is (A). The equation is the same as $\lvert x + 3 \rvert = c + 2$ . With $c = -3$ the right side is $-1$ , and an absolute value is never negative. Choice (B) gives $\lvert x + 3 \rvert = 0$ , which has the one solution $x = -3$ at the vertex $(-3, -2)$ . Choices (C) and (D) give positive distances, so each has two solutions.

What are all solutions of $\lvert x - 5 \rvert = 2x - 1$ ? (A) $2$ only, (B) $-4$ only, (C) $-4$ and $2$ , (D) there is no solution

Show the answer

The correct choice is (A). The case $x - 5 = 2x - 1$ gives $x = -4$ , but then the left side is $\lvert -9 \rvert = 9$ while the right side is $-9$ , so it fails. The case $x - 5 = -(2x - 1)$ gives $3x = 6$ , so $x = 2$ , and both sides equal $3$ . Choice (B) solves only the first case and skips the check, (C) is what you get by solving both cases without checking, and (D) ignores the valid branch.

Absolute Value Equations and Graphs

Theory, part 65

Practice, part 66

Key terms

Absolute value

Distance

Case split

Piecewise interpretation

Vertex

Branch condition

Check yourself

How this lesson connects

Builds on

Used later in