SOLUTION: Find three consecutive odd integers where the sum of the first and two times the seconds equal to one third of the largest

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Question 1199316: Find three consecutive odd integers where the sum of the first and two times the seconds equal to one third of the largest
Found 3 solutions by josgarithmetic, Theo, MathLover1:
Answer by josgarithmetic(39630)   (Show Source):
You can put this solution on YOUR website!
odd integers

Description,

ANSWER:

Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
x = the first odd integer
x + 2 = the second odd integer
x + 4 = the third odd integer.

the statement says that

Find three consecutive odd integers where the sum of the first and two times the seconds equal to one third of the largest

the equation for that is:
x + 2 * (x + 2) = 1/3 * (x + 4)
simplify to get:
x + 2x + 4 = 1/3 * (x + 4)
simplify to get:
3x + 4 = 1/3 * (x + 4)
multiply both sides of this equation by 3 to get:
3 * (3x + 4) = x + 4
simplify to get:
9x + 12 = x + 4
subtract x from both sides of the equation and subtract 12 from both sides of this equation to get:
8x = -8
solve for x to get:
x = -1
that's your first consecutive odd integer.
your first 3 consecutive odd integers are -1, 1, 3
the problem states that sum of the first and two times the seconds equal to one third of the largest
that becomes -1 + 2 * 1 = 1/3 * 3
simplify to get:
-1 + 2 = 1
combine like terms to get:
1 = 1
this confirms the solution is correct.
your solution is that the 3 consecutive odd integers are -1, 1, 3.

Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!

Thus the three consecutive odd integers are:
, and

if the sum of the first and two times the seconds equal to one third of the largest, we have

.............multiply by

and

three consecutive odd integers are:
,, and