Question

Simplify. Consider all cases.

f

14−(6−x)
(14-(6-x) is surround by two absolute bars)

Answer 1

Explanation:

As (14-(6-x) is surrounded by two absolute bars.

so

`\left|14-\left(6-x\right)\right|`

Expanding
`14-\left(6-x\right)`

`=14-6+x`

`\mathrm{Subtract\:the\:numbers:}\:14-6=8`

`=x+8`

so

`=\left|x+8\right|`

Thus

`\left|14-\left(6-x\right)\right|=\left|x+8\right|`

It can also be solved by x, such as

`\left|x+8\right|=0`

`\mathrm{Apply\:absolute\:rule}:\quad \mathrm{If}\:|u|\:=\:0\:\mathrm{then}\:u\:=\:0`

`x=-8`