Step-by-step explanation:
As (14-(6-x) is surrounded by two absolute bars.
so
[tex]\left|14-\left(6-x\right)\right|[/tex]
Expanding [tex]14-\left(6-x\right)[/tex]
[tex]=14-6+x[/tex]
[tex]\mathrm{Subtract\:the\:numbers:}\:14-6=8[/tex]
[tex]=x+8[/tex]
so
[tex]=\left|x+8\right|[/tex]
Thus
[tex]\left|14-\left(6-x\right)\right|=\left|x+8\right|[/tex]
It can also be solved by x, such as
[tex]\left|x+8\right|=0[/tex]
[tex]\mathrm{Apply\:absolute\:rule}:\quad \mathrm{If}\:|u|\:=\:0\:\mathrm{then}\:u\:=\:0[/tex]
[tex]x=-8[/tex]
