THEOREM
solving absolute value sentences
Let
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$\,x\in\mathbb{R}\,$,
and let
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$\,k\ge 0\,$.
Then,
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$$
\begin{gather}
x = k\ \ \text{ is equivalent to }\ \ x = \pm k \\
\\
x \lt k\ \ \ \text{ is equivalent to }\ \ k \lt x \lt k \\
x \le k\ \ \ \text{ is equivalent to }\ \ k \le x \le k \\
\\
x \gt k\ \ \ \text{ is equivalent to }\ \ x\lt k\ \ \text{ or }\ \ x\gt k \\
x \ge k\ \ \ \text{ is equivalent to }\ \ x\le k\ \ \text{ or }\ \ x\ge k
\end{gather}
$$
EXAMPLE (an absolute value equation):
Solve:
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$2  3x = 7$
Solution:
Write a nice, clean list of equivalent sentences:
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$2  3x = 7$

(original equation) 
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$23x = \pm 7$ 
(check that $\,k\ge 0\,$; use the theorem) 
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$23x = 7\ \text{ or }\ 23x = 7$ 
(expand the plus/minus) 
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$3x = 5\ \text{ or }\ 3x = 9$ 
(subtract $\,2\,$ from both sides of both equations) 
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$\displaystyle
x = \frac{5}{3}\ \text{ or } x = 3$ 
(divide both sides of both equations by $\,3\,$) 
It's a good idea to check your solutions:
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$2  3(\frac{5}{3})\ \overset{\text{?}}{=}\ 7$,
$2 + 5 = 7$ Check!
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$2  3(3)\ \overset{\text{?}}{=}\ 7$,
$2  9 = 7$ Check!
EXAMPLE (an absolute value inequality involving ‘less than’):
Solve:
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$36x + 7 \le 9$
Solution:
To use the theorem, you must have the absolute value
all by itself
on one side of the equation.
Thus, your first job is to
isolate the absolute value:
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$36x + 7 \le 9$

(original sentence) 
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$6x + 7 \le 3$

(divide both sides by $\,3$) 
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$3 \le 6x + 7 \le 3$

(check that $\,k \ge 0\,$; use the theorem) 
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$10 \le 6x \le 4$

(subtract $\,7\,$ from all three parts of the compound inequality) 
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$\displaystyle \frac{10}{6} \ge x \ge \frac{4}{6}$

(divide all three parts by $\,6\,$; change direction of inequality symbols) 
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$\displaystyle \frac{2}{3} \le x \le \frac{5}{3}$

(simplify fractions; write in the conventional way) 
Check the ‘boundaries’ of the solution set:
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$36(\frac{2}{3}) + 7 = 34 + 7 = 33 = 9$ Check!
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$36(\frac{5}{3}) + 7 = 310 + 7 = 33 = 9$ Check!
EXAMPLE (an absolute value inequality involving ‘greater than’):
Solve:
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$36x + 7 \ge 9$
Solution:
To use the theorem, you must have the absolute value
all by itself
on one side of the equation.
Thus, your first job is to
isolate the absolute value:
[beautiful math coming... please be patient]
$36x + 7 \ge 9$

(original sentence) 
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$6x + 7 \ge 3$

(divide both sides by $\,3$) 
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$6x + 7 \le 3\ \ \text{or}\ \ 6x + 7\ge 3$

(check that $\,k \ge 0\,$; use the theorem) 
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$6x\le 10\ \ \text{or}\ \ 6x\ge 4$

(subtract $\,7\,$ from both sides of both subsentences) 
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$\displaystyle x\ge\frac{10}{6}\ \ \text{or}\ \ x\le \frac{4}{6}$

(divide by $\,6\,$; change direction of inequality symbols) 
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$\displaystyle x\ge\frac{5}{3}\ \ \text{or}\ \ x\le \frac{2}{3}$

(simplify fractions) 
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$\displaystyle x\le \frac{2}{3}\ \ \text{or}\ \ x\ge\frac{5}{3}$

(in the web exercise, the ‘less than’ part is always reported first) 
Check the ‘boundaries’ of the solution set:
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$36(\frac{2}{3}) + 7 = 34 + 7 = 33 = 9$ Check!
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$36(\frac{5}{3}) + 7 = 310 + 7 = 33 = 9$ Check!
EXAMPLE (an absolute value equation that is always false):
Solve:
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$3x + 1 = 5$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
Can absolute value ever be negative? NO!
No matter what number you substitute for $\,x\,$,
the lefthand side of the equation
will always be a number that is greater than or equal to zero.
Therefore, this sentence has no solutions. It is always false.
EXAMPLE (an absolute value inequality that is always false):
Solve:
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$5  2x \lt 3$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
Can absolute value ever be negative? NO!
No matter what number you substitute for $\,x\,$,
the lefthand side of the inequality
will always be a number that is greater than or equal to zero,
so it can't possibly be less than $\,3\,$.
Therefore, this sentence has no solutions. It is always false.
EXAMPLE (an absolute value inequality that is always true):
Solve:
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$5  2x \gt 3$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
No matter what number you substitute for $\,x\,$,
the lefthand side of the inequality
will always be a number that is greater than or equal to zero,
so it will always be greater than $\,3\,$.
Therefore, this sentence has all real numbers as solutions. It is always true.
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Identifying Perfect Squares
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
Solve the given absolute value sentence.
Write the result in the most conventional way.
For more advanced students, a graph is displayed.
For example, the inequality $\,2  3x \lt 7\,$
is optionally accompanied by the
graph of $\,y = 2  3x\,$ (the left side of the inequality, dashed green)
and the graph of
$\,y = 7\,$ (the right side of the inequality, solid purple).
In this example, you are finding the values of $\,x\,$ where the green
graph lies below the purple graph.
Click the “show/hide graph” button if you prefer not to see the graph.