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Equations and inequalities containing
absolute value
absolute value
The absolute value of a positive number is the number itself, the absolute value of \latex{ 0 } is \latex{ 0 }, and the absolute value of a negative number is the opposite of the number.
Example 1
Let us solve the following equation and inequalities.
- \latex{\left|x-1\right|=2}
- \latex{\left|x-1\right|\lt 2}
- \latex{\left|x-1\right|\gt 2}
Solution (a)
During the solution it is enough to apply the definition. Equation a) is obviously satisfied if
\latex{ x - 1 = 2 } or \latex{ x – 1 = -2. }
The solutions of the two set-up equations: \latex{ x_1 = 3 } and \latex{ x_2 = -1. }
Solution (b)
Inequality b) will be true for those numbers for which:
\latex{ -2 \lt x - 1 \lt 2. }
It means we have to solve two inequalities. One of them is:
\latex{ -2 \lt x - 1 }, therefore \latex{ –1 \lt x. }
The other one is:
x – 1 < 2, therefore x < 3.
Thus the solutions of inequality b) are those numbers for which \latex{ -1 \lt x \lt 3 }, i.e. the real numbers on interval \latex{ ] -1; 3[. } (Figure 13)
Solution (c)
In this case we can act similarly, but here we can use the results of exercises a) and b), and thus the missing real numbers are: \latex{ x \lt -1 \, \text{or} \, 3 \lt x. } The solution in the form of intervals is as follows: \latex{\left]-\infty ;-1\left[\cup \right]3;\infty \right[ }.
Figure 13
\latex{-1\lt x}
\latex{x\lt 3}
\latex{ 3 }
\latex{ -1 }
Example 2
Let us solve the following equation and inequality.
- \latex{\left|x+3\right|-2=3x}
- \latex{\left|x+3\right|-2\lt 3x}
Solution (a)
In case a) we first give a graphical solution. For this as first step let us rearrange the equation to the following form:
\latex{\left|x+3\right|=3x+2}
Then let us plot the graphs of the functions on the two sides of the equation, and let us find their intersection point in the figure. (Figure 14)
In figure 14 it can be seen that the two graphs intersect each other at the value \latex{x=\frac{1}{2}}.
This will also be the solution of the equation; we can make sure of its correctness when checking.
⯁ ⯁ ⯁
Figure 14
\latex{y=\left|x+3\right| }
\latex{y=3x+2}
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ x }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ -5 }
\latex{ -1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ y }
We can also solve the equation in an algebraic way. As first step we interpret the absolute value:
\latex{\left|x+3\right|=\begin{cases} x+3, \; if\; x+3\geq 0, i.e. \;x\geq-3,\\ -x-3, \; if\; x+3\lt0, i.e.\; x\lt-3.\end{cases}}
Thus the exercise can be split into two cases according to the ranges in which we are looking for the solutions:
- If \latex{x\geq -3}, then the equation is the following (Figure 15):
\latex{ x + 3 - 2 = 3x. }
After rearranging and solving it:
\latex{ x + 1 = 3x, }
\latex{ 1 = 2x, }
\latex{x=\frac{1}{2}}
\latex{ 1 = 2x, }
\latex{x=\frac{1}{2}}
The resulting number is also a root of the original equation since it is in the specified range.
Figure 15
\latex{\frac{1}{2} }
\latex{0}
\latex{-3}
- If \latex{ x \lt -3 }, then (Figure 16):
\latex{-x-3-2=3x},
\latex{-x-5=3x},
\latex{-5=4x}
\latex{x=-\frac{5}{4} }.
\latex{-x-5=3x},
\latex{-5=4x}
\latex{x=-\frac{5}{4} }.
This value however does not meet the condition, thus it will not be a root of the original equation either.
Thus the equation has only one solution: \latex{x=\frac{1}{2} }.
Figure 16
\latex{-\frac{5}{4} }
\latex{0}
\latex{-3}
Solution (b)
When solving the inequality we can use the graphical solution of the equation. First, again rearranging yields the following:
\latex{\left|x+3\right|\lt 3x+2}
After that based on the position of the graphs we can state that in range \latex{x\gt \frac{1}{2} } the graph of the function on the left side is below the graph of the function on the right side, therefore the solution of the inequality will be the interval \latex{\left]\frac{1}{2};\infty \right[ }. (Figure 17)
Figure 17
\latex{y=\left|x+3\right| }
\latex{y=3x+2 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ x }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ -5 }
\latex{ -1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ y }
Example 3
Let us solve the equation \latex{\left|x+1\right| -\left|2x-1\right| =x}.
Solution
Let us interpret the absolute value expressions in the equation.
\latex{ |x+1| = \begin{cases}\color{#faf7f7}- \color{black}x+1, \quad\text{if } x\geq-1; \\ -x-1, \quad\text{if } x\lt-1.\end{cases} \quad |2x-1| = \begin{cases} \color{#faf7f7}-\color{black}2x-1, \quad\text{if } x\geq\frac 1 2; \\ -2x+1, \quad\text{if } x\lt\frac 1 2.\end{cases}}
Based on the resulting domains and according to figure \latex{ 18 } we divide the number line into three ranges.
Figure 18
\latex{x\lt -1}
\latex{-1\leq x\lt \frac{1}{2} }
\latex{\frac{1}{2}\leq x }
\latex{-1}
\latex{\frac{1}{2}}
I.
II.
III.
If \latex{x\lt -1}, then
If \latex{-1\leq x\lt \frac{1}{2} }, then
If \latex{\frac{1}{2}\leq x}, then
\latex{-x-1-\left(-2x+1\right)=x},
\latex{-x-1+2x-1=x},
\latex{-2=0}.
\latex{-x-1+2x-1=x},
\latex{-2=0}.
\latex{x+1-\left(-2x+1\right)=x},
\latex{x+1+2x-1=x},
\latex{3x=x},
\latex{2x=0},
\latex{x=0}.
\latex{x+1+2x-1=x},
\latex{3x=x},
\latex{2x=0},
\latex{x=0}.
\latex{x+1-\left(2x-1\right)=x},
\latex{x+1-2x+1=x},
\latex{2=2x},
\latex{x=1}.
\latex{x+1-2x+1=x},
\latex{2=2x},
\latex{x=1}.
The equation leads to a contradiction, so there is no solution in this range.
The resulting root is also a solution of the original equation since it is an element of the range designated in case II.
The resulting root will also be a solution since it meets condition III.
So the equation will have two solutions: \latex{ x_1 = 0 } and \latex{ x_2 = 1 }\latex{ }, which are also justified by checking.
Example 4
Let us solve the inequality \latex{\left|x+1\right| -\left|2x-1\right| \gt x}.
Solution
During the solution let us use the interpretations of absolute values from the previous exercise. According to this we have to distinguish three cases:
I.
II.
III.
If \latex{x\lt -1}, then
If \latex{-1\leq x\lt \frac{1}{2} }, then
If \latex{\frac{1}{2}\leq x}, then
\latex{-x-1-\left(-2x+1\right)\gt x},
\latex{-x-1+2x-1\gt x},
\latex{-2\gt 0}.
\latex{-x-1+2x-1\gt x},
\latex{-2\gt 0}.
\latex{x+1-\left(-2x+1\right)\gt x},
\latex{x+1+2x-1\gt x},
\latex{3x\gt x},
\latex{2x\gt 0},
\latex{x\gt 0}.
\latex{x+1+2x-1\gt x},
\latex{3x\gt x},
\latex{2x\gt 0},
\latex{x\gt 0}.
\latex{x+1-\left(2x-1\right)\gt x},
\latex{x+1-2x+1\gt x},
\latex{2\gt 2x},
\latex{1\gt x}.
\latex{x+1-2x+1\gt x},
\latex{2\gt 2x},
\latex{1\gt x}.
The equation leads to a contradiction, so there is no solution in this range.
Combining the resulting condition with the condition given at the interpretation we get the following solution:
Taking into consideration the conditions given at the interpretation too, we get the following solution:
Figure 19
Figure 20
\latex{0\lt x\lt \frac{1}{2} }.
\latex{\frac{1}{2}\leq x\lt1 }.
\latex{-1}
\latex{0}
\latex{1}
\latex{\frac{1}{2}}
\latex{-1}
\latex{0}
\latex{\frac{1}{2}}
\latex{1}
Summarised: the solution of the inequality: \latex{ 0 \lt x \lt 1 } which can be represented on a number line:
Figure 21
\latex{ \frac{1}{2} }
\latex{ 0 }
\latex{ 1 }
Example 5
Let us solve the following equation and inequality.
- \latex{\left|\left|x-1\right|-2 \right| -3=0}
- \latex{\left|\left|x-1\right|-2 \right| -3\geq 0}
Solution (a)
The given equation differs from the previous types in containing the absolute value of absolute value expressions.
After rearranging the equation based on the definition:
\latex{\left|\left|x-1\right|-2\right|-3=0},
\latex{\left|x-1\right|-2=3}
\latex{\left|x-1\right|-2=-3}.
In the first case we get that:
\latex{\left|x-1\right|=5}, it implies that
\latex{x-1=5} or \latex{x-1=-5}.
\latex{x-1=5} or \latex{x-1=-5}.
In this case even two solutions are resulting: \latex{ x_1 } = 6 and \latex{ x_2 }\latex{ = -4 }.
In the other case after rearranging the equation \latex{\left|x-1\right|=-1} is resulting which naturally cannot be fulfilled by any real number when taking the range of the absolute value into consideration.
Thus the solution of the exercise is: \latex{ x_1 = 6 } and \latex{ x_2 } \latex{ = -4. } When substituting back into the equation we can make sure that the resulting roots meet the conditions of the exercise.
Solution (b)
It is practical to solve inequality b) graphically. We plot the graph of the function on the left side following the below steps:
\latex{f_{1} \left(x\right) =x-1, f_{2} \left(x\right)=\left|x-1\right| ,f_{3} \left(x\right)=\left|x-1\right| -2,}
\latex{f_{4} \left(x\right)=\left|\left|x-1\right|-2 \right| , f_{5} \left(x\right)=\left|\left|x-1\right|-2 \right| -3}.
\latex{f_{4} \left(x\right)=\left|\left|x-1\right|-2 \right| , f_{5} \left(x\right)=\left|\left|x-1\right|-2 \right| -3}.
After plotting the graphs we get figures \latex{ 22/a–e }.
Figure 22/a
\latex{y=x-1}
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ x }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ y }
Figure 22/b
\latex{y=\left|x-1\right| }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ x }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ y }
Figure 22/c
\latex{y=\left|x-1\right| -2}
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ x }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ y }
Figure 22/d
\latex{y=\left|\left|x-1\right| -2\right| }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ 6 }
\latex{ 7 }
\latex{ 8 }
\latex{ x }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ -5 }
\latex{ -6 }
\latex{ -7 }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ 6 }
\latex{ y }
Figure 22/e
\latex{y=\left|\left|x-1\right| -2\right|-3}
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ 6 }
\latex{ 7 }
\latex{ 8 }
\latex{ x }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ -5 }
\latex{ -6 }
\latex{ -7 }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -4 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ 6 }
\latex{ y }
Based on the last figure we can also see the solution set of the inequality, which consists of the real numbers for which the following condition is met:
\latex{x\leq -4} or \latex{6\leq x}.
The solution in another form: \latex{\left]-\infty ;-4\right] \cup \left[6;\infty \right[}.
The graphical solution also justifies the correctness of the solution of exercise a).
The graphical solution also justifies the correctness of the solution of exercise a).
Exercises
{{exercise_number}}. Solve the following equations on the set of real numbers.
- \latex{\left|x\right| -1=2}
- \latex{\left|x-1\right|=2 }
- \latex{\left|x\right| -1\lt 2}
- \latex{\left|x-1\right| \gt 2}
{{exercise_number}}. Solve the following equations on the set of whole numbers.
- \latex{2x=\left|-x\right| -1}
- \latex{\left|3x-2\right| -1=2x+1}
- \latex{\left|-x+1\right| =\left|x-1\right| }
- \latex{\left|-x-1\right| =-x+2}
{{exercise_number}}. For which real numbers are the following inequalities satisfied?
- \latex{\left|x\right| \geq x}
- \latex{\left|-x\right| \lt x}
- \latex{2+x\lt \left|x-2\right| }
- \latex{\left|2x-4\right| \geq x-1}
{{exercise_number}}. Solve the following equations on the set of whole numbers.
- \latex{\left|x+1\right| +\left|x-1\right| =2}
- \latex{\left|2x-3\right| -\left|x-1\right| =x}
- \latex{\left|\left|x\right| -1\right| -2=3}
- \latex{\left|\left|\left|x-1\right|-2 \right| -3\right| =1}
{{exercise_number}}. Solve the following inequalities on the set of real numbers.
- \latex{\left|x\right| -x+1\lt 0}
- \latex{\left|x\right| +\left|x-1\right| \geq 2}
- \latex{\left|\left|x\right| -1\right| -2\lt3}
- \latex{\left|\left|\left|x-1\right|-2 \right| -3\right| \geq1}