SOLUTION: Find three consecutive (not even) integers such that the sum of the twice the first, six times the second and three times the third is 133.

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Question 123689: Find three consecutive (not even) integers such that the sum of the twice the first, six times the second and three times the third is 133.
Found 2 solutions by checkley71, nadinadan:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
x, x+1, x+2 are the integers.
2x+6(x+1)+3(x+2)=133
2x+6x+6+3x+6=133
11x=133-12
11x=121
x=121
11x=11 answer for the first integer.
11=1=12
11+2=13
proof:
2*11+6*12+3*13=133
22+72+39=133
133=133

Answer by nadinadan(32) About Me  (Show Source):
You can put this solution on YOUR website!
Let's think about the consecutive numbers as A, A+1 and A+2.
Let's replace the words of the problem into mathematical language.
So 2A+6(A+1)+3(A+2)=133
So we simplify these: 2A+6A+6+3A+6=133, so 11A=133-12 so A=11 and A+1=11+1=12 and A+2=11+2=13
So these are the numbers 11,12,13.