Absolute Value Equations

Example 1. Solve the following equation: $$${\left|{x}\right|}=5$$$.

We know, that absolute value makes any number positive.

What are numbers, that when made positive will give $$$5$$$?

They are $$$5$$$ and $$$-5$$$.

So, $$$5$$$ and $$$-5$$$ are the roots of the equation.

Example 2. Solve the following equation: $$${\left|{x}+{3}\right|}={7}$$$.

Absolute value is already isolated, so we have 2 cases:

1. $$${x}+{3}={7}$$$, which gives $$${x}={4}$$$.
2. $$${x}+{3}=-{7}$$$, which gives $$${x}=-{10}$$$.

So, $$${4}$$$ and $$$-{10}$$$ are roots of the equation.

Example 3. Solve the following equation: $$${\left|{2}{x}+{5}\right|}-{8}={14}$$$.

First, isolate absolute value.

For this add $$${8}$$$ to both sides of the equation: $$${\left|{2}{x}+{5}\right|}={22}$$$.

Absolute value is isolated, so we have 2 cases:

1. $$${2}{x}+{5}={22}$$$, which gives $$${x}=\frac{{17}}{{2}}$$$.
2. $$${2}{x}+{5}=-{22}$$$, which gives $$${x}=-\frac{{27}}{{2}}$$$.

So, $$$\frac{{17}}{{2}}$$$ and $$$-\frac{{27}}{{2}}$$$ are roots of the equation.

Example 4. Solve the equation $$${\left|{2}{x}+{3}\right|}={5}-{x}$$$.

We can use above rule, but carefully.

Indeed, rule requires the right side to be positive number, but we don't know whether $$${5}-{x}$$$ is positive or negative.

So, after finding potential roots, we need to check them.

Thus, we have two options:

1. $$${2}{x}+{3}={5}-{x}$$$, which gives $$${x}=\frac{{2}}{{3}}$$$.
2. $$${2}{x}+{3}=-{\left({5}-{x}\right)}$$$, which gives $$${x}=-{8}$$$.

We've got two potential solutions, now need to check them.

1. $$${\left|{2}\cdot\frac{{2}}{{3}}+{3}\right|}={5}-\frac{{2}}{{3}}$$$

   $$${\left|\frac{{7}}{{3}}\right|}=\frac{{7}}{{3}}$$$

   $$$\frac{{7}}{{3}}=\frac{{7}}{{3}}$$$ (OK)

2. $$${\left|{2}\cdot{\left(-{8}\right)}+{3}\right|}={5}-{\left(-{8}\right)}$$$

   $$${\left|-{13}\right|}={5}+{8}$$$

   $$${13}={13}$$$ (OK)

So, $$$\frac{{2}}{{3}}$$$ and $$$-{8}$$$ are roots of the equation.

Example 5. Solve the equation $$${\left|{x}+{5}\right|}={2}{x}+{1}$$$.

As in example 4 we can use above rule, but carefully.

Indeed, rule requires the right side to be positive number, but we don't know whether $$${2}{x}+{1}$$$ is positive or negative.

So, after finding potential roots, we need to check them.

Thus, we have two options:

$$${x}+{5}={2}{x}+{1}$$$, which gives $$${x}={4}$$$.

$$${x}+{5}=-{\left({2}{x}+{1}\right)}$$$, which gives $$${x}=-{2}$$$.

We've got two potential solutions, now need to check them.

1. $$${\left|{4}+{5}\right|}={2}\cdot{4}+{1}$$$

   $$${\left|{9}\right|}={9}$$$

   $$${9}={9}$$$ (OK)

2. $$${\left|-{2}+{5}\right|}={2}\cdot{\left(-{2}\right)}+{1}$$$

   $$${\left|{3}\right|}=-{3}$$$

   $$${3}\ne-{3}$$$ (NOT OK)

So, the only root of the equation is $$${4}$$$.

Example 6. Solve the equation $$${\left|{2}{x}+{3}\right|}={\left|{3}{x}+{9}\right|}$$$.

It seems, that above rule won't work here. But we don't need it here.

What is absolute value? It makes any number positive.

So, absolute value expressions are equal, when quantities inside absolute value bars are equal or equal with different signs.

We again have two cases here:

1. $$${2}{x}+{3}={3}{x}+{9}$$$, which gives $$${x}=-{6}$$$
2. $$${2}{x}+{3}=-{\left({3}{x}+{9}\right)}$$$, which gives $$${x}=-\frac{{12}}{{5}}$$$

Thus, equation has two roots: $$$-{6}$$$ and $$$-\frac{{12}}{{5}}$$$.