Question 744978
Find three consecutive odd integers, so that 3 times their sum is 5 more than 8 times the middle one.


x = 1st odd integer
x + 2 = 2nd odd integer
x + 4 = 3rd odd integer   {odd and also even integers increase by 2}


3(x + x + 2 + x + 4) = 8(x + 2) + 5   {sum is equal to 5 more than 8 times the middle one}
3(3x + 6) = 8x + 16 + 5   {combined like terms on left, used distributive property on right}
9x + 18 = 8x + 21   {combined like terms}
x = 3   {subtracted 8x and subtracted 18 from each side}
x + 2 = 5   {substituted 3, in for x, into x + 2}
x + 4 = 7   {substituted 3, in for x, into x + 4}


3,5, and 7 are the three consecutive odd integers
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