Modulus equations (Absolute Value Equalities): Difficult Problems with Solutions

Solution:
Notice that [tex]|4-2x|=2|x-2|[/tex] and [tex]|6-3x|=3|x-2|[/tex]:
[tex]|x-2|-2|x-2|-3|x-2|=-8 \Rightarrow -4|x-2|=-8 \Rightarrow |x-2|=2[/tex]
[tex]x-2=2 \Rightarrow x=4[/tex]; [tex]x-2=-2 \Rightarrow x=0[/tex]

Solution:
Two absolute values are equal when the expressions inside are equal or opposite:
[tex]2-3x=9+4x \Rightarrow -7x=7 \Rightarrow x=-1[/tex]
[tex]2-3x=-(9+4x) \Rightarrow 2-3x=-9-4x \Rightarrow x=-11[/tex]
[tex]-1+(-11)=-12[/tex]

Solution:
The expressions under the absolute values change sign at [tex]x=0[/tex] and [tex]x=\frac{1}{2}[/tex]. We consider three intervals:
[tex]x\lt 0:\ -x-(2x-1)=3 \Rightarrow x=-\frac{2}{3}[/tex] (it is in the interval)
[tex]0\le x\lt \frac{1}{2}:\ x-(2x-1)=3 \Rightarrow x=-2[/tex] (not in the interval)
[tex]x\ge \frac{1}{2}:\ x+2x-1=3 \Rightarrow x=\frac{4}{3}[/tex] (it is in the interval)

Solution:
For [tex]x\lt 3[/tex]: [tex]3-x>0[/tex] and [tex]x-5\lt 0[/tex], so [tex]|3-x|=3-x[/tex] and [tex]|x-5|=5-x[/tex]:
[tex]3-x-x(5-x)=3 \Rightarrow x^2-6x=0 \Rightarrow x(x-6)=0[/tex]
[tex]x=0[/tex] or [tex]x=6[/tex]; only [tex]x=0[/tex] satisfies [tex]x\lt 3[/tex].

Solution:
[tex]a\lt 0 \Rightarrow a-1\lt 0 \Rightarrow |a-1|=1-a[/tex]
[tex]\big|a-2-(1-a)\big|=|2a-3|[/tex]
[tex]a\lt 0 \Rightarrow 2a-3\lt 0 \Rightarrow |2a-3|=3-2a[/tex]

Solution:
[tex]a\lt 0[/tex]: an absolute value is never negative – no solutions.
[tex]a=0[/tex]: [tex]2x-1=0 \Rightarrow x=\frac{1}{2}[/tex] – one solution.
[tex]a>0[/tex]: [tex]2x-1=a[/tex] or [tex]2x-1=-a[/tex], i.e. [tex]x=\frac{1+a}{2}[/tex] or [tex]x=\frac{1-a}{2}[/tex] – two solutions.

Solution:
[tex]x_1=m+3[/tex], [tex]x_2=m-3[/tex]
[tex](m+3)(m-3)\lt 0[/tex]
The product of two numbers is negative when they have opposite signs; since [tex]m+3>m-3[/tex], we need [tex]m+3>0[/tex] and [tex]m-3\lt 0[/tex]:
[tex]-3\lt m\lt 3[/tex]

Solution:
An absolute value cannot be negative, so there is no solution exactly when the right side is negative:
[tex]m^2-4m+3=(m-1)(m-3)\lt 0[/tex]
The factors have opposite signs; since [tex]m-1>m-3[/tex]: [tex]m-1>0[/tex] and [tex]m-3\lt 0[/tex], i.e. [tex]1\lt m\lt 3[/tex].

Solution:
[tex](x-4)^2=(2x+1)^2 \Rightarrow x-4=2x+1[/tex] or [tex]x-4=-(2x+1)[/tex]
[tex]x=-5[/tex] or [tex]x=1[/tex]
[tex]|x-m|=3 \Rightarrow x=m-3[/tex] or [tex]x=m+3[/tex]
The roots must coincide: [tex]m-3=-5[/tex] and [tex]m+3=1[/tex], so [tex]m=-2[/tex].
[tex]m^2=4[/tex]

Solution:
A sum of two absolute values (non-negative numbers) is [tex]0[/tex] only when both are [tex]0[/tex]:
[tex]x^2-6x+8=(x-2)(x-4)=0 \Rightarrow x=2[/tex] or [tex]x=4[/tex]
[tex]x^2-6x+5=(x-1)(x-5)=0 \Rightarrow x=1[/tex] or [tex]x=5[/tex]
There is no common value, so the equation has no solution.

Solution:
Since [tex]x^2-4x=(x-2)^2-4[/tex] and [tex](x-2)^2=|x-2|^2[/tex], let [tex]t=|x-2|\ge 0[/tex]:
[tex]t^2-4+t=8 \Rightarrow t^2+t-12=0 \Rightarrow (t+4)(t-3)=0[/tex]
[tex]t=-4[/tex] is impossible ([tex]t\ge 0[/tex]), so [tex]t=3[/tex]: [tex]|x-2|=3 \Rightarrow x=5[/tex] or [tex]x=-1[/tex].