SOLUTION: how can i find 5 consecutive integers such that the sum of the 4th and the 5th is 56 more than triple the 1st?

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Question 942592: how can i find 5 consecutive integers such that the sum of the 4th and the 5th is 56 more than triple the 1st?
Found 2 solutions by josmiceli, macston:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the consecutive integers be:
n, n+1, n+2, n+3, and n+4
-------------------------
+n+%2B+3+%2B+n+%2B+4+=+3n+%2B+56+
+2n+%2B+7+=+3n+%2B+56+
+n+=+7+-+56+
+n+=+-49+
+n+%2B+1+=+-48+
+n+%2B+2+=+-47+
+n+%2B+3+=+-46+
+n+%2B+4+=+-45+
note that integers can be positive or negative
----------------
The consecutive integers are -49, -48, -47, -46, and -45
check:
+n+%2B+3+%2B+n+%2B+4+=+3n+%2B+56+
+-46+-+45+=+3%2A%28+-49+%29+%2B+56+
+-91+=+-147+%2B+56+
+-91+=+-91+

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
first=x, second =x+1, third=x+2, fourth=x+3, fifth=x+4
3x+56=x+3+x+4
3x+56=2x+7 subtract 2x from each side
3x-2x+56=2x+7-2x
x+56=7 subtract 56 from each side
x+56-56=7-56
x=-49 first= -49
fourth=x+3=-46
fifth=x+4=-45
CHECK: Sum offourth and fifth is 56 more than triple first
-45+-46=3(-49)+56
-91=(-147)+56
-91=-91