Question 228196
The sum of three consecutive integers is 47 less than the least of the integers.  Show work to understand.


Step 1.  Let n be the first and the least of the integers.


Step 2.  Let n+1 and n+2 be the next two consecutive integers.


Step 3.  Let n+n+1+n+2 be the sum of the three consecutive integers.


Step 4.  Let n-47 be 47 less than the least of the integers


Step 5.  Then n+n+1+n+2=n-47 since the sum of three consecutive integers is 47 less than the least of the integers


Step 6.  Solving n+n+1+n+2=n-47 leads to the following steps


{{{3n+3=n-47}}}


Subtract 3 from both sides of the equation


{{{3n+3-3=n-47-3}}}


{{{3n=n-50}}}


Subtract n from both sides of the equation


{{{3n-n=n-50-n}}}


{{{2n=-50}}}


Divide by 2 to both sides of the equation


{{{2n/2=-50/2}}}


{{{n=-25}}}  {{{n+1=-24}}} and {{{n+2=-23}}}


Check if sum if 47 less than the least.


{{{-25-24-23=-25-47}}}


{{{-72=-72}}} which is a true statement


Step 7.  ANSWER:  The three consecutive numbers are -25, -24, and -23.


I hope the above steps were helpful.


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Good luck in your studies!


Respectfully,
Dr J