SOLUTION: Find three consecutive even integers such that the sum of the twice the first, six times the second and three times the third is 133.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Find three consecutive even integers such that the sum of the twice the first, six times the second and three times the third is 133.      Log On


   



Question 123685: Find three consecutive even integers such that the sum of the twice the first, six times the second and three times the third is 133.
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
x, x+2, x+4
2x+6(x+2)+3(x+4)=133
2x+6x+12+3x+12=133
11x=133-24
11x=109
x=109/11
x=9.9090909 not an even number.
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ARE YOU SURE YOU HAVE THE PROPER WORDING IN THIS PROBLEM??????
THREE EVEN NUMBERS WHEN ADDED ADD OR MULTIPLIED MUST BE AN EVEN NUMBER 133 IS NOT AN EVEN NUMBER.
IF THESE ARE NOT EVEN NINTERGERS THEN:
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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