#### mcallendar

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Where would I put the parenthesis in this problem.

6+3/3+2-1-3=3

6+3/3+2-1-3=3

- Thread starter mcallendar
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Where would I put the parenthesis in this problem.

6+3/3+2-1-3=3

6+3/3+2-1-3=3

- Joined
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Please follow the rules of posting in this forum - enunciated at:Where would I put the parenthesis in this problem.

6+3/3+2-1-3=3

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/

Please share your work/thoughts with us - so that we know where to begin to help you.

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Presumably you are asking where you can put parentheses in thatWhere would I put the parenthesis in this problem.

6+3/3+2-1-3=3

All you can do is to try something; there is no magic or formula for this sort of thing. It's just meant to give you practice in evaluating, and perhaps to discover some general principles.

What have you tried?

I will tell you that I haven't yet seen an answer, so it may either be impossible, or use some unexpected trick. (Or maybe I just missed something obvious.)

Where would I put the parenthesis in this problem.

6+3/3+2-1-3=3

It's been more than a week. I'm posting my full solutions for people to consider.

\(\displaystyle 6(+ \ 3/3)(+ \ 2 - 1) - 3 \ = \ 3\)

\(\displaystyle 6+ (3/3)(+ \ 2 - 1)( - 3) \ = \ 3\)

\(\displaystyle 6(+ \ (3/3)(+ \ 2) - 1) - 3 \ = \ 3\)

\(\displaystyle (6(+ \ 3)/3)(+ \ 2 - 1) - 3 \ = \ 3\)

\(\displaystyle 6 + ((3/3)(+ \ 2) - 1)(-3) \ = \ 3\)

\(\displaystyle 6(+ \ 3/(3(+ \ 2 - 1))) - 3 \ = 3 \)

\(\displaystyle 6 + (3/(3(+ \ 2 - 1)))(-3) \ = \ 3\)

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