How do you solve #3(n-1)-2=4(n-4)-(n-11)#?

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Answer 1 · 2015-12-09T04:55:46

There are infinite solutions.

#3(n-1)-2=4(n-4)-(n-11)#

Distribute the #3# on the left side.

#3n-3-2=4(n-4)-(n-11)#

Distribute the #4# on the right side.

#3n-3-2=4n-16-(n-11)#

Distribute the #-1# on the right side.

#3n-3-2=4n-16-n+11#

Combine like terms on both sides.

#3n-5=4n-n-16+11#

Simplify.

#3n-5=3n-5#

There infinite solutions.