SOLUTION: Find three consecutive odd integers such that the sum of the first two plus three times the largest is 99

Question 919612: Find three consecutive odd integers such that the sum of the first two plus three times the largest is 99
Answer by algebrapro18(249) (Show Source): You can put this solution on YOUR website!
Lets let n = the first of our odd numbers. Then notice that the odd numbers increase by 2. So the next consecutive odd number would be n+2. And the third consecutive odd number would be n+2+2 = n+4.

Now we know from what we're trying to find that the sum of the first and second consecutive odd number plus 3 times the third odd number is 99. So writing this as an equation we get:

n+n+2+3(n+4)=99

Now we can solve that equation:

n+n+2+3(n+4)=99 Distribute the 3
n+n+2+3n+12=99 Combine like terms
5n+14=99 Subtract 14 from both sides
5n = 85 divide both sides by 5
n = 17

Now we know that our first odd number is 17. Plugging back into our expressions for the second and third consecutive odd numbers we get 17+2 and 17+4 or 19 and 21 respectively.

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