How do you simplify #abs((-(3-1)+5)div(-2))#?

2 Answers

#3/2#

Explanation:

The absolute value of a number is always positive, but before we can do the absolute value, we need to simplify what is inside of it. So let's do that:

#abs((-(3-1)+5)-:(-2))#

The first thing I'm seeing is that it's a bit difficult to see what is going on, so I'll add some colour and help things a bit. Specifically, I'm going to colour a term that is within brackets that needs to be done before we can so the division (which is the operation that jumped out at me at first!):

#abs(color(green)((-(3-1)+5))-:(-2))#

So I'm going to do the green section separately, then put the result back in:

#color(green)((-(3-1)+5))=(-(2)+5)=(-2+5)=(3)#

and now let's put the solution to this part back into the original:

#abs(3-:(-2))=abs(-3/2)#

and now we can take the absolute value:

#abs(-3/2)=3/2#

Jan 14, 2017

#3/2#

Explanation:

#color(blue)("Introduction to concept")#

The | | turns what ever is inside it into a positive value.
AS an example: #|-7|=|+7|=7#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider: "color(green)((-color(red)((3-1))+5)#

Bracket within brackets is called nested brackets. In this situation you consider each pair of brackets working outwards

Step 1: #" "color(red)((3-1) = +2)#

Step 2: #" "color(green)((-color(red)((3-1))+5)" "->" "color(green)((-color(red)((+2))+5)#

#" "->" "(-2+5) = (+3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Putting it all together")#

#|color(white)(.)(-(3-1)+5)-:(-2)color(white)(.)|" " =" " |color(white)(.)3-:(-2)color(white)(.)|#

#" "=" "|color(white)(.)-3/2color(white)(.)|#

#" "=" "+3/2#