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

Solution:
[tex]0\times x=-5[/tex] has no solution, so [tex](a-5)x=2+a[/tex] must also have no solution: [tex]a-5=0[/tex] and [tex]2+a\neq 0[/tex], i.e. [tex]a=5[/tex].
For [tex]x\lt -2[/tex]: [tex]|x+2|=-x-2[/tex] and [tex]|x|=-x[/tex]:
[tex]-x-2-x=-x+5 \Rightarrow -x=7 \Rightarrow x=-7[/tex]
[tex]x=-7\lt -2[/tex], so it is a solution.

Solution:
From [tex]|b|\,c=4>0[/tex] and [tex]|b|\ge 0[/tex] it follows that [tex]c>0[/tex].
Then [tex]|2a+1|\,c\ge 0[/tex] and [tex]|b+1|\ge 0[/tex]; their sum is [tex]0[/tex] only if both are [tex]0[/tex]:
[tex]|2a+1|=0 \Rightarrow a=-\frac{1}{2}[/tex]; [tex]|b+1|=0 \Rightarrow b=-1[/tex], [tex]c=4[/tex]

Solution:
[tex]f(1)=g(1) \Rightarrow 1+a=2+b \Rightarrow a-b=1[/tex]
[tex]B(0, f(0))=B(0, 2+a)[/tex], [tex]C(0, g(0))=C(0, b)[/tex]
[tex]BC=|2+a-b|=|2+1|=3[/tex]
The height from [tex]A[/tex] to [tex]BC[/tex] (the [tex]y[/tex]-axis) equals the abscissa of [tex]A[/tex], i.e. [tex]1[/tex]:
[tex]S=\frac{3\times 1}{2}=\frac{3}{2}[/tex]
Remarkably, the area does not depend on [tex]a[/tex] and [tex]b[/tex].

Solution:
Case [tex]x\ge 5[/tex]: [tex]x-5=bx \Rightarrow x(1-b)=5[/tex]. For [tex]b\neq 1[/tex], [tex]x=\frac{5}{1-b}[/tex], and [tex]x\ge 5[/tex] holds exactly when [tex]0\lt 1-b\le 1[/tex], i.e. [tex]0\le b\lt 1[/tex].
Case [tex]x\lt 5[/tex]: [tex]5-x=bx \Rightarrow x(1+b)=5[/tex]. For [tex]b\neq -1[/tex], [tex]x=\frac{5}{1+b}[/tex], and [tex]x\lt 5[/tex] holds exactly when [tex]1+b>1[/tex] or [tex]1+b\lt 0[/tex], i.e. [tex]b>0[/tex] or [tex]b\lt -1[/tex].
So there is a solution when [tex]b\ge 0[/tex] or [tex]b\lt -1[/tex], and no solution when [tex]-1\le b\lt 0[/tex].