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Equalities and inequalities with absolute value

Absolute value equalities and inequalities are equalities and inequalities in which the unknown or unknowns are in the aboslute value. Through the examples beneath we examine and explain how to solve these types of problems. All the definitons and properties used are described here.

Example 1.

Calculate the value of the given expression if: $\ a = 2$, $\ b = -5$, $\ |(|a|-|b|)| = ?$.

First, the variables $a$ and $b$ are replced with their value: $\ |(|2|-|-5|)| = ?$

If the expression contains more than one absolute value they are treated the same way parenthesis are, the first to be solved are the ones inside the absolute value then those outside of it.

$\ |(|2|-|-5|)| = |2-5| = |-3| = 3$

 

Example 2.

Calculate the value of $\displaystyle{\frac{2|a – 3b|}{|b|} }$ if: $ a = -\displaystyle{\frac{-1}{2}}$ , $b = \displaystyle{\frac{4}{3}}$.

$\displaystyle{\frac{2 \mid -\frac{1}{2} – 4 \mid}{\mid \frac{4}{3} \mid}=\frac{2 \mid -\frac{9}{2} \mid}{\frac{4}{3}} = \frac{9}{\frac{4}{3}} = \frac{27}{4}}$

 

Example 3.

Solve for x: $|x| + 2 = 4$.

The unknowns are first moved, whether they are bound with a module or not, to the left side, and the constants to the right.

$\ |x| = 2$.

Since absolute value is defined as the distance of a number from the origin on the number line, these equation has two solutions, $2$ and $-2$, because both are the same distance away from the origin, or $\ |2| = |-2| = 2$.

 

Example 4.

$\mid -0.2x \mid + \left | -\displaystyle{\frac{3}{4}} \right | = 2$

$\left | -\displaystyle{\frac{1}{5}x} \right | + \displaystyle{\frac{3}{4} = 2}$

$\left | -\displaystyle{\frac{1}{5}x} \right | = \displaystyle{\frac{5}{4}}$

Now we can use the property $ \mid a * b \mid = \mid a \mid * \mid b \mid$

$\displaystyle{\left | -\frac{1}{5} \right | \mid x \mid = \frac{5}{4}}$

$\displaystyle{\frac{1}{5} \mid x \mid = \frac{5}{4} / \cdot 5}$

$\displaystyle{\mid x \mid = \frac{25}{4}} $

$x_1 = \displaystyle{\frac{25}{4}}, x_2 = -\displaystyle{\frac{25}{4}}$

 

Example 5.

$|-2x – 2| = 2$

$|x|=0$ iff $x=0$, however $|-2x – 2|=0$ for some other $x$. When this $x$ is known, then for every $x$ on one side of the number line $|-2x – 2|$ is negative and for every $x$ on the other side is positive. How does one find these particular $x$ for which $|-2x – 2|=0$? The process is called “searching zeroes”. The expression is equated with 0 and then, if $y=-2x – 2$, then $|y|=0$ iff $y=0$ which is the same as $-2x – 2=0$.

$\ -2x – 2 = 0$

$\ -2x = 2$

$\ x = -1$

Iff $x=-1$, then $|-2x – 2|=0$. When the value of $x$ is less, the expression has one sign, and if $x$ is greater than the other sign. How does one know on which side of this point will the values be positive and on which negative?

It is enough to look at any random number from interval $\ <-\infty, -1 >$ and see the value of the module in that case.

For $x=-2$, which is less then $x=-1$, the expression is equal to $2$.

$|-2 \cdot{(-2)} – 2| = 2$

$|4 – 2| = 2$

$|2|=2$

The value inside the module is greater than zero so for the interval $\ <-\infty, -1 >$ everything inside the module will remain the same, and for the interval $\ <-1,\infty >$ the expression inside the module changes the sign.

$\ 2x + 2 = 2 \Rightarrow \ x = 0$.

 

Example 6.

$\displaystyle{\left|\frac{x-1}{x-2}\right|}=\displaystyle{\frac{1}{2}}$

 $|a|=|b| \Leftrightarrow$ $a=b$ or $a=-b$

$\displaystyle{\frac{|x-1|}{|x-2|}}=\displaystyle{\frac{1}{2}}$ $/\cdot 2, |x-2|, x \neq 2$

$2|x-1|=|x-2|$

$|2x-2|=|x-2|$

The first case: $2x-2=x-2$.

$2x-2=x-2   /+(-x+2)$

$2x-2-x+2=x-2-x+2$

$x=0$

The second case: $2x-2=-x+2$.

$2x-2=-x+2 /+(x-2)$

$2x-2+x-2=-x+2+x-2$

$x=\displaystyle{\frac{4}{3}}$

Example 7.

$\ |x| > 2$

Variable $x$ can be either positive or negative, so two different cases are studied:

Case 1  $\ x > 0$ which gives $\ x > 2$.

Case 2  $\ x < 0$ which gives $\ – x > 2$ and $\ x < – 2$.

The final solution is the union of those two intervals: $\ <-\infty, -2 > \cup < 2, \infty >$.

 

Example 8.

$\ |x + 3| > 4$

The first step is the same as for equalities.

$\ x + 3 = 0 \Rightarrow \ x = – 3$

For $x = 0$, $\ 0 + 3 = 3 > 0$.

  1. case: $\ x > – 3 \Rightarrow$ everything inside module remains the same $ \Rightarrow \ x + 3 > 4$ , $\ x > 1$
  2. case: $\ x < – 3 \Rightarrow$ everything inside module changes sign $\Rightarrow \ – x – 3 > 4$ , $\ – x > 7$ , $\ x < – 7$

In the first case, the solutions must satisfy both inequalities: $\ x >- 3$ and $x > 1$, so the solution is their intersection $\ <1, \infty>$.

In the second case, the solutions must satisfy both inequalities: $\ x < – 3$ and $x < -7$, so the solution is their intersection $\ <\infty, -7>$

 

The final solution is the union between the first and the second case $\ <\infty, -7> \cup<1, \infty>$.