## Equations

In Section 1.5, we learned that
$\bbox[#F2F2F2,5px,border:2px solid black]{\large |x|=c\geq 0\qquad \Leftrightarrow \qquad x=\pm c}$
provided $c\geq0$. So to solve equations involving an absolute value follow these steps:

1. Isolate the absolute value expression on one side and the rest of terms on the other side. That is, rewrite the equation as
$|P|=Q$ where $P$ and $Q$ are two expressions in $x$ [to indicate the dependence on $x$, we may write them as $P(x)$ and $Q(x)$].

2. Equate the expression inside the absolute value notation once with + the quantity on the other side and once with – the quantity on the other side.
$P=Q\qquad\text{or}\qquad P=-Q$
3. Solve both equations.
• When $Q<0$, the equation will not have a solution because always $|P|\geq0$. When $Q$ is an expression, we need to substitute the solutions in $Q$ to make sure that $Q\geq0$.
Example 1
Solve each equation:
(a) $|2x-3|=5$
(b) $|5x-7|+9=0.$

Solution
(a)
$|2x-3|=5\quad\Leftrightarrow\quad2x-3=\pm5$ \begin{align*}
2x-3 & =5\Rightarrow x=4\\
2x-3 & =-5\Rightarrow x=-1
\end{align*}
So the solutions are $x=4$ and $x=-1$. We can check to see these values satisfy the equation, but it is not necessary because the right-hand side is a positive number.
(b)
$|5x-7|=-9$ because always $|5x-7|\geq0$ and $-9<0$, this equation does not have a solution.
Example 2
Solve each equation:
(a) $|3-2x|+5x=18$
(b) $|4x+3|+3x=10$
Solution
(a) We isolate the absolute value expression on one side:

$|3-2x|=18-5x\tag{i}$ which is equivalent to
$3-2x=18-5x\qquad\text{or\}\qquad3-2x=-18+5x}}.$ Solving each equation:
$3-2x=18-5x\Rightarrow3x=15\Rightarrow x=5$ or
$3-2x=-18+5x\Rightarrow7x=21\Rightarrow x=3$ If we substitute $5$ for $x$ in $18-5x$ it becomes $18-25<0$. Because the RHS of (i) is negative and the LHS is nonnegative, $x=5$ cannot be a solution. But if we substitute 3 for $x$ in $18-5x$, it becomes $18-15=3>0$, so the RHS and the LHS of (i) are both nonnegative and $x=3$ is the only solution.

Alternatively, we can substitute $x=5$ and $x=3$ in the original equation and check if they satisfy the equation.

(b) Similar to (a)
$|4x+3|=10-3x\tag{ii}$ $\quad\Leftrightarrow\quad4x+3=\pm(10-3x)$ We have to solve two equations:
$(1)\qquad4x+3=10-3x\Rightarrow7x=7\Rightarrow x=1$ $(2)\qquad4x+3=-10+3x\Rightarrow x=-13$ Substituting $1$ for $x$ in RHS of (ii) ($10-3x$) gives $7$. Because the LHS and RHS of (ii) are nonnegative, $x=1$ is a solution. Substituting $-13$ for $x$ in the RHS of (ii) gives a positive number, so $x=-13$ is another solution. Therefore the solutions are $x=1$ and $x=-13$.

Alternatively we can substitute $x=1$ and $x=-13$ in the original equation and see which one satisfies the equation.

When there are more than one absolute value, for example when we have
$|P|+|R|=Q$ where $P, Q$, and $R$ are some expressions, the above technique may not work. In such cases, we need to find where $P$ and $R$ are positive and where they are negative and then solve the equation in the same way that we solve regular equations.

Example 3
Solve $|2x+4|+4|14-3x|=30$.
Solution
Using the definition of the absolute value

$|2x+4|=\begin{cases} 2x+4 & x\ge-2\\ -(2x+4) & x<-2 \end{cases}$ $|14-3x|=|3x-14|=\begin{cases} 3x-14 & x\ge\frac{14}{3}\approx4.67\\ 14-3x & x<\frac{14}{3} \end{cases}$

When $x\ge14/3\approx4.67$:
\begin{align*}
|2x+4|+4|14-3x|=14x-52 & =30\\
14x-52 & =30\\
14x & =82\\
x & =\frac{82}{14}=\frac{41}{7}\approx5.86
\end{align*}
Because $x=41/7$ lies in the interval $[14/3,\infty)$, it is consistent with our assumption that $x\ge14/3$ and thus $x=41/7$ is a solution.

When $-2\le x<\frac{14}{3}$:
\begin{align*}
|2x+4|+4|14-3x|=-10x+60 & =30\\
-10x+60 & =30\\
-10x & =-30\\
x & =3
\end{align*}
Because $x=3$ lies in the interval $[-2,\frac{14}{3}]$, it is consistent with our assumption that $-2\le x<14/3$ and thus $x=3$ is a solution.

When $x<-2$:
\begin{align*}
|2x+4|+4|14-3x|=-14x+52 & =30\\
-14x+52 & =30\\
-14x & =-22\\
x & =\frac{11}{7}
\end{align*}

Because $x=11/7$ does not lie in the interval $(-\infty,-2)$, it is \uline{not} consistent with our assumption that $x<-2$, and thus $x=11/7$ cannot be a solution. Therefore, the solution set is
$\{3,\frac{41}{7}\}.$ It does not matter in which interval we consider the endpoints because
$\text{at }x=-2\qquad-14x+52=-10x+60=80$ and
$\text{at }x=14/3\qquad-10x+60=14x-52=40/3.$

## Inequalities

To solve absolute value inequalities, recall (see Section 1.4):

1.  $|x|<c$ is equivalent to $-c<x<c$.
2. $|x|>c$ is equivalent $x>c$ or $x<-c$
where $c$ is a positive number.

The above equivalent statements hold true if we replace $<$ by $\le$ and $>$ by $\ge$.

Example 4
Solve each of the following inequalities
(a) $|x-3|<5$
(b) $|5-4x|>6.$
Solution
(a) The inequality $|x-3|<5$ is equivalent to
$-5<x-3<5$

$-2<x<8$ The solution set is the interval $(-2,8)$.
(b) The inequality $|5-4x|>6$ is equivalent to
$5-4x>6\qquad\text{or}\qquad5-4x<-6}}$
$-4x>1\qquad\text{or}\qquad-4x<-11}}$
Divide each term by $-4$:
$x<-\frac{1}{4}\qquad\text{or}\qquad x>\frac{1}{4}$ [For the last step, recall that when we divide both sides of an inequality by a negative number, the direction of the inequality changes.]
$\{x|\ x<-\frac{1}{4}\quad\text{or}\quad x>\frac{1}{4}\}=(-\infty,-\frac{1}{4})\cup(\frac{1}{4},\infty).$