How do you solve #-12= 12r + 24#?

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Answer 1 · 2016-11-16T14:14:06

#r = -3#

Perform the necessary mathematics to isolate #r# while keeping the equation balanced:

#-12 - 24 = 12r + 24 - 24#

#-36 = 12r + 0#

#-36 = 12r#

#-36/12 = (12r)/12#

#-3 = 12/12 r#

#-3 = 1 * r#

#r = -3#