How do you solve #10=3( r + 1) + 1#?

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Answer 1 · 2018-01-16T18:00:30

r=2

#10=3(r+1)+1#
#10=3r+3+1#
#10=3r+4 // -4,#

#6=3r //:3,#

#rArrr=2#

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Answer 2 · 2018-01-16T18:02:21

#r = 2#

#10 = 3(r+1)+1#

subtract #1#:

#9 = 3(r+1)#

divide by #3#:

#3 = r+1#

#r+1 = 3#

subtract #1#:

#r = 2#

alternatively:

expand brackets:
#10 = 3r + 3 + 1#

collect like terms:

#10 = 3r+4#

subtract #4#:

#3r = 6#

divide by #3#:

#r = 2#