How do you solve #n/2+3=n-1#?

Question

338 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-13T02:00:03

#n=7#

#n/2+3=n-1#

Multiply through by 2:

#n+6=2n-2#

Bring terms in #n# to the LHS and constants to the RHS:

#n-2n = -2-6#

#-n=-8#

#n=8#

NB: With a little practice you should be able to perform some of these steps in one go.

To check the answer, replace #n=8# in the original equation:

#8/2+3 =8-1 -> 7=7# Answer is therefore correct.