SOLUTION: HELP!!!! 4 times the third, of three consecutive integers, is 51 less than 3 times the sum of the first and second integers. Find the third integer?

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Question 838995: HELP!!!! 4 times the third, of three consecutive integers, is 51 less than 3 times the sum of the first and second integers.
Find the third integer?

Found 2 solutions by ewatrrr, josmiceli:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
three consecutive integers: highlight_green%28x%29, highlight_green%28x%2B1%29, highlight_green%28x%2B2%29
Question states***
4 times the third is 51 less than 3 times the sum of the first and second integers.
***4(x+2) = 3[x + (x+1)] -51 | Solving for x
4x + 8 = 3(2x+1) - 51
4x + 8 = 6x + 3 - 51
56 = 2x
28 = x , the first Integer. The third integer is highlight%28+30%29
CHECKING our answer***
4%2A30+=+3%2828%2B29%29+-+51
120 = 120
Wish You the Best in your Studies.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the integers be +n+, +n%2B1+, and +n+%2B+2+
+4%2A%28+n%2B2+%29+=+3%2A%28+n+%2B+n+%2B+1+%29+-+51+
+4n+%2B+8+=+3%2A%28+2n+%2B+1+%29+-+51+
+4n+=+6n+%2B+3+-+51+-+8+
+4n+=+6n+-+56+
+6n+-+4n+=+56+
+2n+=+56+
+n+=+28+
+n+%2B+1+=+29+
+n+%2B+2+=+30+
The 3rd integer is 30
check:
+4%2A%28+n%2B2+%29+=+3%2A%28+n+%2B+n+%2B+1+%29+-+51+
+4%2A%28+30+%29+=+3%2A%28+28+%2B+29+%29+-+51+
+120+=+3%2A57+-+51+
+120+=+171+-+51+
+120+=+120+
OK