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How do you find 3 consecutive integers such that their sum is 27 less than three times the smallest?

Answer 1

Eva Adamson

Call the smallest #n#; you have: #n+(n+1)+(n+2)=27-3n# #3n+3=27-3n# #6n=24# #n=4# So: #4+5+6=15=27-(3×4)=27-12=15#

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Answer 2

Abigail Allen

Let the three consecutive integers be ( x ), ( x+1 ), and ( x+2 ).
According to the given condition, the sum of the three integers is equal to ( 3x + (x+1) + (x+2) = 27 - 3x ).
Solving the equation ( 3x + (x+1) + (x+2) = 27 - 3x ) gives:
( 3x + x + 1 + x + 2 = 27 - 3x )
( 5x + 3 = 27 - 3x )
( 5x + 3x = 27 - 3 )
( 8x = 24 )
( x = 3 )
So, the three consecutive integers are ( 3 ), ( 4 ), and ( 5 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.