# What is the solution set for #abs(2x – 3) – 8 = –1#?

##### 1 Answer

#### Answer:

#### Explanation:

Start by isolating the modulus on one side of the equation by adding

#|2x-3| - color(red)(cancel(color(black)(8))) + color(red)(cancel(color(black)(8))) = -1 + 8#

#|2x-3| = 7#

As you know, the absolute value of a real number is always positive **irrespective** of that number's sign.

This tells you that you have two cases to think about, one in which the expression that's inside the modulus is *positive*, and the other in which the expression inside the modulus is *negative*.

#2x-3 >0 implies |2x-3| = 2x-3#

This will make your equation take the form

#2x - 3 = 7#

#2x = 10 implies x= 10/2 = color(green)(5)#

#2x-3<0 implies |2x-3| = -(2x-3)#

This time, you have

#-(2x-3) = 7#

#-2x + 3 = 7#

#-2x = 4 implies x = 4/((-2)) = color(green)(-2)#

So there ar eactually two possible solutions to this equation, one that makes *positive*, *negative*,