# How do you solve #3 < absx – 5#?

##### 1 Answer

#### Answer:

#### Explanation:

Start by isolating the modulus on one side of the inequality. You can do this by adding

#3 + 5 < |x| - color(red)(cancel(color(black)(5))) + color(red)(cancel(color(black)(5)))#

#8 < |x|#

This is of course equivalent to

#|x| > 8#

Now, you need to take into account the fact that *positive* or *negative*, which means that you get

#x>0 implies |x| = x#

For positive values of

#x > 8#

#x<0 implies |x| = -x#

For negative values of

#-x > 8#

Multiply both sides by *do not forget* that the sign of the inequality changes when you multiply or divide by a *negative number*

#-1 * (-x) color(red)(<) 8 * (-1)#

#x < -8#

This means that your origininal inequality will be tru for any value of *smaller* than *bigger* than