How do you solve #–2(x – 3) ≥ 5 – (x + 3)#?

2 Answers
Alan P.
Mar 7, 2018

#x <= 4#

Explanation:

Given
#color(white)("XXX")-2(x-3) >= 5-(x+3)#

Simplifying each side independently
#color(white)("XXX")-2x+6 >= 2-x#

We could add #2x# to both sides (you can add any amount to both sides of an inequality without effecting its validity or direction)
#color(white)("XXX")6 >=2+x#

Then, if we subtract #2# from both sides
#color(white)("XXX")4 >= x#...or, if you like your variables on the left
#color(white)("XXX")x <= 4#

F. Javier B.
Mar 7, 2018

See details below

Explanation:

First expand inequation removing braquets

#-2x+6>=5-x-3#. Now, adding to both sides in order to transpose terms, we have

#6-5+3>= -x+2x# this is the same to

#4>=x#

The inequation has as solutions all numbers #x# lower than #4#, it's say #(-oo,4]#

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