Jesse thinks of a number.
He then does the following operations:
Multiply by 3, subtract 7, multiply by 3 then add 45 (in that order).
He finds that the number he ends up with is 13 times his original number.
What was Jesse's original number?
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This question is best answered by forming an algebraic equation then solving it. Let Jesse's original number be x.
First operation gives 3x
Second operation gives 3x- 7
Third operation gives 3(3x- 7)
Fourth operation gives 3(3x- 7) + 45
This is equal to 13 times the original number
3(3x - 7) + 45 = 13x
9x- 21 + 45 = 13x
x = 6
Jesse's original number was 6.
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