The modulus, or absolute value, measures distance from $0$ on the number line.
Therefore, absolute value is always non-negative:
$$|x|\geq 0.$$
For a positive constant $a$, the following relationships are fundamental.
When the boundary on the right-hand side contains $x$, extra care is needed.
For example, in an equation such as $|A(x)|=B(x)$, the right-hand side must satisfy
$$B(x)\geq 0,$$
because an absolute value can never be negative.
For equations of the form
$$|A(x)|=B(x),$$
use the two possible cases:
$$A(x)=B(x)\quad\text{or}\quad A(x)=-B(x),$$
but always check the resulting answers in the original equation.
EXAMPLE 2
Solve the equation
$$|x-1|=5x-10.$$
Solution:
First note that the right-hand side must be non-negative:
$$5x-10\geq 0\quad\Longrightarrow\quad x\geq 2.$$
Now solve the two algebraic cases:
$$\begin{aligned}
x-1&=5x-10
&&\quad\text{or}\quad&
x-1&=-(5x-10)\\
x-1&=5x-10
&&&
x-1&=-5x+10\\
9&=4x
&&&
6x&=11\\
x&=\dfrac{9}{4}
&&&
x&=\dfrac{11}{6}.
\end{aligned}$$
Check the restriction $x\geq 2$. The value $\dfrac{9}{4}$ is valid, but
$\dfrac{11}{6}$ is rejected because $\dfrac{11}{6}\lt 2$.
Therefore,
$$x=\dfrac{9}{4}.$$
EXAMPLE 3
Solve the inequality
$$|x-1|\lt 5x-10.$$
Solution:
Since the left-hand side is non-negative, the right-hand side must be positive:
$$5x-10\gt 0\quad\Longrightarrow\quad x\gt 2.$$
For $B(x)\gt 0$, the inequality $|A(x)|\lt B(x)$ becomes
$$-B(x)\lt A(x)\lt B(x).$$
Here this gives
$$-(5x-10)\lt x-1\lt 5x-10.$$
When an expression contains one or more absolute values, the most reliable method is
to split the number line using the zeros of the expressions inside the absolute value signs.
On each interval, every absolute value can be replaced by either a positive or negative version
of its inside expression.
The definition
$$|u|=
\begin{cases}
u, & u\geq 0,\\
-u, & u\lt 0
\end{cases}$$
is the basis of the piecewise method.
EXAMPLE 4
Solve
$$|1-x|=3x+2.$$
Solution:
The expression inside the modulus is zero when
$$1-x=0\quad\Longrightarrow\quad x=1.$$
So the number line is split into two intervals:
$$x\lt 1\qquad\text{and}\qquad x\geq 1.$$
Case 1: $x\lt 1$. Then $1-x\gt 0$, so $|1-x|=1-x$.
$$\begin{aligned}
1-x&=3x+2\\
-4x&=1\\
x&=-\dfrac{1}{4}.
\end{aligned}$$
Since $-\dfrac{1}{4}\lt 1$, this solution is accepted.
Case 2: $x\geq 1$. Then $1-x\leq 0$, so $|1-x|=x-1$.
$$\begin{aligned}
x-1&=3x+2\\
-2x&=3\\
x&=-\dfrac{3}{2}.
\end{aligned}$$
Since $-\dfrac{3}{2}$ does not satisfy $x\geq 1$, this solution is rejected.
Therefore,
$$x=-\dfrac{1}{4}.$$
EXAMPLE 5
Solve
$$|x-1|+|x-2|=x.$$
Solution:
The zeros of the absolute value expressions are
$$x=1\qquad\text{and}\qquad x=2.$$
These split the real line into three intervals:
$$x\lt 1,\qquad 1\leq x\lt 2,\qquad x\geq 2.$$
Interval
$x\lt 1$
$1\leq x\lt 2$
$x\geq 2$
$|x-1|$
$-(x-1)=1-x$
$x-1$
$x-1$
$|x-2|$
$-(x-2)=2-x$
$-(x-2)=2-x$
$x-2$
Case 1: $x\lt 1$.
$$\begin{aligned}
(1-x)+(2-x)&=x\\
3-2x&=x\\
3&=3x\\
x&=1.
\end{aligned}$$
Since $1$ does not satisfy $x\lt 1$, this is rejected.
Case 2: $1\leq x\lt 2$.
$$\begin{aligned}
(x-1)+(2-x)&=x\\
1&=x.
\end{aligned}$$
Since $x=1$ satisfies $1\leq x\lt 2$, this is accepted.
Case 3: $x\geq 2$.
$$\begin{aligned}
(x-1)+(x-2)&=x\\
2x-3&=x\\
x&=3.
\end{aligned}$$
Since $3\geq 2$, this is accepted.
Therefore,
$$x=1\quad\text{or}\quad x=3.$$
4. Graphing Linear Modulus Expressions
A sum of linear modulus expressions produces a piecewise linear graph.
The graph changes direction only at the zeroes of the expressions inside the absolute values.
EXAMPLE 6 (Graphing Linear Modulus Sequences)
Sketch the graph of
$$f(x)=|x-1|+|x-2|-x.$$
Solution:
The breakpoints are
$$x=1\qquad\text{and}\qquad x=2.$$