SOLUTION: Find three consecutive even integers such that the sum of three times the first integer, one-third the second integer, and eight times the third integer is -58

Question 28820: Find three consecutive even integers such that the sum of three times the first integer, one-third the second integer, and eight times the third integer is -58
Answer by xcentaur(357) (Show Source): You can put this solution on YOUR website!
let any integer be x
to make it definitely even,make it 2x
now three consecutive even integers would mean 2x,(2x+2),(2x+4)

we have been given a condition,
"sum of three times the first integer, one-third the second integer, and eight times the third integer is -58"
i've colored them different so you can follow it in the working
3(2x)+1/3(2x+2)+8(2x+4)=-58
[6x] + [2/3(x+1)] + [2/8(x+2)] = -58
6x + 2/3(x+1) + 1/4(x+2) = -58
multiply throughout by 3
18x + 2(x+1) + 3/4(x+2) = -3(58)
multiply throughout by 4
72x + 8(x+1) + 3(x+2) = -3(4)(58)
72x + 8x + 8 + 3x + 6 = -12(58)
83x + 14 = -696
83x = -696-14= -710
x= -710/83

Now this value of x is not only negative,it is also not an integral value.
Clearly this '-58' cannot be correct. please check your question and post again.

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