How do you solve #abs(-2r-1)=11#?

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Answer 1 · 2017-10-30T17:54:09

#r = -6, r = 5#

This is an absolute value equation meaning that we need to split this into two equations; the original equation and the opposite of it:

#-2r-1 = 11# and #-2r-1=-11#
Notice that the #11# became #-11#.

Now we can solve this by simplifying. Let's simplify the first one:
#-2r-1 = 11#
#-2r = 12#
#r = -6#

Second one:
#-2r-1 = -11#
#-2r = -10#
#r = 5#

Now let's check our work by plugging what we got for r back into the original equation #|-2r-1| = 11#.
#|-2(-6) -1|= 11#
#|12-1| = 11#
#11 = 11# :)

#|-2(5)-1|=11#
#|-10-1| = 11#
#|-11| = 11#
#11 = 11# :)

So our two solutions are #r = -6# and #r = 5#.