Modulus equations (Absolute Value Equalities): Problems with Solutions

Solution:
[tex]x^2=|x|^2[/tex], so [tex]x^2|x|=|x|^2|x|=|x|^3[/tex] for every [tex]x[/tex].
The other three are false for negative [tex]x[/tex]; for example, for [tex]x=-1[/tex]:
[tex]|x|^3=1, x|x|^2=-1, x^3=-1, |x^3|=1[/tex]

Solution:
[tex]|x|\ge 0[/tex], so [tex]5-|x|\le 5[/tex]; the value [tex]5[/tex] is reached for [tex]x=0[/tex].

Solution:
[tex]A=5\times |-3|-2\times |5|-4\div |-2|=5\times 3-2\times 5-4\div 2=15-10-2=3[/tex]

Solution:
[tex]2x-1=3[/tex] or [tex]2x-1=-3[/tex]
[tex]x=2[/tex] or [tex]x=-1[/tex]

Solution:
[tex]2x-3=7 \Rightarrow x=5[/tex]
[tex]2x-3=-7 \Rightarrow x=-2[/tex]
[tex]5+(-2)=3[/tex]

Solution:
[tex]4=4|3x-1| \Rightarrow |3x-1|=1[/tex]
[tex]3x-1=1 \Rightarrow x=\frac{2}{3}[/tex]; [tex]3x-1=-1 \Rightarrow x=0[/tex]

Solution:
The left side is non-negative, so [tex]1-3x\ge 0[/tex], i.e. [tex]x\le \frac{1}{3}[/tex].
For such [tex]x[/tex]: [tex]3x-2\lt 0[/tex], so [tex]|3x-2|=2-3x[/tex], and the equation becomes [tex]2-3x=1-3x \Rightarrow 2=1[/tex] – impossible. The equation has no solution.

Solution:
[tex]|-5x-1|=|5x+1|=16[/tex]
[tex]5x+1=16 \Rightarrow x=3[/tex]; [tex]5x+1=-16 \Rightarrow x=-\frac{17}{5}=-3. 4[/tex]
Integers between [tex]-3. 4[/tex] and [tex]3[/tex]:
[tex]-3-2-1+0+1+2=-3[/tex]

Solution:
For [tex]a\lt -4[/tex]: [tex]a\lt 0[/tex] and [tex]a+4\lt 0[/tex], so [tex]|a|=-a[/tex] and [tex]|a+4|=-a-4[/tex].
[tex]3|a|+|a+4|=-3a-a-4=-4a-4[/tex]

Solution:
For [tex]2\lt x\lt 3[/tex]: [tex]x>0[/tex], [tex]3-x>0[/tex], [tex]6-3x\lt 0[/tex], so
[tex]2x-(3-x)-(3x-6)=2x-3+x-3x+6=3[/tex]

Solution:
Domain: $x+2\ge0 \Leftrightarrow x \ge -2, x\in[-2,+\infin)$

$|x|=x+2$

Case 1:
$x=x+2$
$0 \times x=2$
There is no solution.


Case 2
$x=-(x+2)$
$x=-x-2$
$2x=-2$
$x=-1\in [-2,+\infin)$

Answer: $-1$

Solution:
$2|x+1|+1=3$
$2|x+1|=2$
$|x+1|=1$

Case 1:
$x+1=1$
$x=0$

Case 2:
$x+1=-1$
$x=-2$

Answer: $-2, 0$

Solution:
$|x^2-5x|=6$

Case 1:
$x^2-5x=6$
$x^2-5x-6=0$
$D=49$
$x_1=-1$
$x_2=6$


Case 2:
$x^2-5x=-6$
$x^2-5x+6=0$
$D=1$
$x_1=2$
$x_2=3$

Answer: $-1, 2, 3, 6$

Solution:
$|4-|x||=2$

Case 1:

$4-|x|=2$
$-|x|=-2$
$|x|=2$
$x=2$ or $x=-2$

Case 2:

$4-|x|=-2$
$-|x|=-6$
$|x|=6$
$x=6$ or $x=-6$

Answer: $-6, -2, 2, 6$

Solution:

	$(-\infin,1)$	$[1, 2)$	$[2,+\infin)$
$x-1$	-	+	+
$x-2$	-	-	+

Case 1: $x\in(-\infin,1)$

$|x-1|+|x-2|=1$
$-(x-1)+(-(x-2))=1$
$-x+1-x+2=1$
$-2x=-2$
$x=1 \notin (-\infin,1)$ There is no solution in this interval.


Case 2: $x\in [1, 2)$

$|x-1|+|x-2|=1$
$(x-1)+(-(x-2))=1$
$0x=0 \Rightarrow \forall x \in [1, 2)$


Case 3: $x\in [2,+\infin)$

$|x-1|+|x-2|=1$
$(x-1)+(x-2)=1$
$2x=4$
$x=2\in [2,+\infin)$

Answer: $[1,2]$

Solution:
$2x-1=0\Leftrightarrow 2x=1 \Leftrightarrow x=\frac{1}{2}$
$3x+2=0\Leftrightarrow 3x=-2 \Leftrightarrow x=-\frac{2}{3}$
$x=0$

	$(-\infin,-\frac{2}{3})$	$[-\frac{2}{3}, 0)$	$[0,\frac{1}{2})$	$[\frac{1}{2},+\infin)$
$2x-1$	-	-	-	+
$3x+2$	-	+	+	+
$x$	-	-	+	+

Case 1 $x\in(-\infin,-\frac{2}{3})$
$|2x-1|=-(2x-1)$
$|3x+2|=-(3x+2)$
$|x|=-x$

$|2x-1|+|3x+2|=3-|x|$
$-(2x-1)-(3x+2)=3-(-x)$
$-2x+1-3x-2=3+x$
$-5x-1=3+x$
$-6x=4$
$x=-\frac{2}{3} \notin (-\infin,-\frac{2}{3}) \Rightarrow -\frac{2}{3}$ is not a solution


Case 2 $x\in[-\frac{2}{3}, 0)$
$|2x-1|=-(2x-1)$
$|3x+2|=(3x+2)$
$|x|=-x$

$-(2x-1)+(3x+2)=3-(-x)$
$-2x+1+3x+2=3+x$
$x+3=3+x$
$0\cdot x=0 \Leftrightarrow \forall x \in [-\frac{2}{3}, 0)$


Case 3 $x\in[0, \frac{1}{2})$
$|2x-1|=-(2x-1)$
$|3x+2|=(3x+2)$
$|x|=x$

$-(2x-1)+(3x+2)=3-x$
$-2x+1+3x+2=3-x$
$x+3=3-x$
$2x+3=0$
$x=0 \in [0, \frac{1}{2}) \Rightarrow $ is a solution


Case 4 $x\in[\frac{1}{2}, +\infin)$
$|2x-1|=(2x-1)$
$|3x+2|=(3x+2)$
$|x|=x$

$(2x-1)+(3x+2)=3-x$
$2x-1+3x+2=3-x$
$5x+1=3-x$
$6x=2$
$x=\frac{1}{3} \notin [\frac{1}{2}, +\infin) \Rightarrow \frac{1}{3}$ is not a solution


Answer: $x \in [-\frac{2}{3}, 0]$

Solution:
$2x+3 = 0 \Leftrightarrow 2x=-3 \Leftrightarrow x=-\frac{3}{2}$

$2x^2-x-1=0$
$D=0$
$x_1=1, x_2=-\frac{1}{2}$

	$(-\infin,-\frac{3}{2})$	$[-\frac{3}{2}, -\frac{1}{2})$	$[-\frac{1}{2},1)$	$[1,+\infin)$
$2x+3$	-	+	+	+
$2x^2-x-1$	+	+	-	+

Case 1 $x \in (-\infin,-\frac{3}{2})$
$|2x+3| = -(2x+3)$
$|2x^2-x-1|=2x^2-x-1$

$-(2x+3)=(2x^2-x-1)+1$
$-2x-3=2x^2-x-1+1$
$-2x-3=2x^2-x$
$-2x^2-x-3=0$
$2x^2+x+3=0$
$D < 0, x\in \varnothing$

Case 2 $x \in [-\frac{3}{2},-\frac{1}{2})$
$|2x+3| = 2x+3$
$|2x^2-x-1|=2x^2-x-1$

$2x+3=2x^2-x-1+1$
$2x+3-2x^2+x=0$
$-2x^2+3x+3=0$
$2x^2-3x-3=0$
$D = 33$
$x_{1,2}=\frac{3\pm\sqrt{33}}{4}$
$x_1=\frac{3+\sqrt{33}}{4} \approx 2.9 \notin [-\frac{3}{2},-\frac{1}{2})$ - not a solution
$x_1=\frac{3-\sqrt{33}}{4} \approx -0.69 \in [-\frac{3}{2},-\frac{1}{2})$ - solution

Case 3 $x \in [-\frac{1}{2},1)$
$|2x+3| = 2x+3$
$|2x^2-x-1|= -(2x^2-x-1)$

$2x+3=-(2x^2-x-1)+1$
$2x+3=-2x^2+x+1+1$
$2x^2+2x-x+3-2=0$
$2x^2+x+1=0$
$D < 0 \Rightarrow x\in \varnothing$

Case 4 $x \in [1,+\infin)$
$|2x+3| = 2x+3$
$|2x^2-x-1|=2x^2-x-1$

$2x+3=2x^2-x-1+1$
$2x+3-2x^2+x=0$
$-2x^2+3x+3=0$
$2x^2-3x-3=0$
$D = 33$
$x_{1,2}=\frac{3\pm\sqrt{33}}{4}$
$x_1=\frac{3+\sqrt{33}}{4} \approx 2.9 \in [1,+\infin)$ - solution
$x_1=\frac{3-\sqrt{33}}{4} \approx -0.69 \notin [1,+\infin)$ - not a solution

Answer: $\frac{3\pm\sqrt{33}}{4}$

Solution:
$2-x=0 \Leftrightarrow x=2$
$3x-1=0 \Leftrightarrow x=\frac{1}{3}$
$x-4=0 \Leftrightarrow x=4$

	$(-\infin,\frac{1}{3})$	$[\frac{1}{3}, 2)$	$[2,4)$	$[4,+\infin)$
$2-x$	+	+	-	-
$3x-1$	-	+	+	+
$x-4$	-	-	-	+

Case 1 $x\in (-\infin, \frac{1}{3})$

$|2-x|=2-x$
$|3x-1|=-(3x-1)$
$|x-4|=-(x-4)$

$(2-x)-(3x-1)=5x+2(x-4)$
$2-x-3x+1=5x+2x-8$
$-4x+3=7x-8$
$-4x-7x=-8-3$
$-11x=-11$
$x=1 \notin (-\infin, \frac{1}{3}) \Rightarrow $ not a solution

Case 2 $x\in [\frac{1}{3},2)$

$|2-x|=(2-x)$
$|3x-1|=(3x-1)$
$|x-4|=-(x-4)$

$(2-x)+(3x-1)=5x+2(x-4)$
$2-x+3x-1=5x+2x-8$
$2x+1=7x-8$
$2x-7x=-8-1$
$-5x=-9$
$5x=9$
$x=\frac{9}{5} \in [\frac{1}{3},2) \Rightarrow $ is a solution

Case 3 $x\in [2, 4)$

$|2-x|=-(2-x)$
$|3x-1|=(3x-1)$
$|x-4|=-(x-4)$

$-(2-x)+(3x-1)=5x+2(x-4)$
$-2+x+3x-1=5x+2x-8$
$4x-3=7x-8$
$4x-7x=-8+3$
$-3x=-5$
$3x=5$
$x=\frac{5}{3} \notin [2, 4) \Rightarrow $ is not a solution

Case 4 $x\in [4, +\infin)$

$|2-x|=-(2-x)$
$|3x-1|=(3x-1)$
$|x-4|=(x-4)$

$-(2-x)+(3x-1)=5x-2(x-4)$
$-2+x+3x-1=5x-2x+8$
$4x-3=3x+8$
$4x-3x=8+3$
$x=11 \in [4, +\infin) \Rightarrow $ is a solution

Answer: $\frac{9}{5}, 11$