Question 515864: i dont know how to write these two equations( find four consecutive integers such that the sum of the two greater is 17 less than twice the sum of the two smallest)and (find four consecutive odd integers such that the sum of the two greatest is four times the smallest
Found 2 solutions by drcole, MathTherapy:
Answer by drcole(72)   (Show Source):
You can put this solution on YOUR website! Let x be the first consecutive integer. Then x + 1 is the second consecutive integer, x + 2 is the third, and x + 3 is the fourth and largest.
Let's look at the first equation. The sum of the two greater consecutive integers is:
(x + 2) + (x + 3)
2x + 5 (combining like terms)
The sum of the two smaller consecutive integers is:
x + (x + 1)
2x + 1 (combining like terms)
We know that the sum of the two greater integers is 17 less than the twice the sum of the two smaller integers. Writing this as an equation, we get:
(Sum of Two Greater Integers) = 2(Sum of Two Smaller Integers) - 17
(2x + 5) = 2(2x + 1) - 17
2x + 5 = 4x + 2 - 17 (distributing 2 on the right side)
2x + 5 = 4x - 15 (combining like terms)
2x + 20 = 4x (add 15 to both sides)
20 = 2x (subtract 2x from both sides)
10 = x
So the smallest integer is 10, making the other integers 11, 12, and 13. Let's check this: the sum of the two greater integers is 12 + 13 = 25. The sum of the two smaller integers is 10 + 11 = 21. 17 less than twice the sum of the two smaller integers is 2(21) - 17 = 42 - 17 = 25, the same as the sum of the two greater integers. So these integers work.
Let's now do the second equation. Again, the sum of the two greater integers is:
(x + 2) + (x + 3) = 2x + 5
Four times the smallest consecutive integer would be 4x. We know that the sum of the two greater integers is four times the smallest integer, and we can now write this fact as an algebraic equation and solve for x:
2x + 5 = 4x
5 = 2x (subtract 2x from both sides)
2.5 = x (divide both sides by 2)
So we got that the smaller integer is 2.5, but that cannot be right, since 2.5 is not an integer. So this problem doesn't have a solution. Actually, this makes sense: the sum of the two greater integers would be the sum of an even integer and an odd integer, which is odd, but four times the smallest integer would be an even integer, so they cannot be equal to each other. Hence the second problem has no solution.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
i dont know how to write these two equations( find four consecutive integers such that the sum of the two greater is 17 less than twice the sum of the two smallest)and (find four consecutive odd integers such that the sum of the two greatest is four times the smallest
I.
Let the smallest integer be S
Then the others are: S + 1, S + 2, and S + 3
We therefore have: S + 2 + S + 3 = 2(S + S + 1) - 17
2S + 5 = 2(2S + 1) - 17
2S + 5 = 4S + 2 - 17
2S - 4S = - 15 - 5
- 2S = - 20
S, or smallest integer =  , or 
The 3 other numbers are:  ,  , and 
II
find four consecutive odd integers such that the sum of the two greatest is four times the smallest
Let the smallest odd integer be S
Then the others are: S + 2, S + 4, and S + 6
We therefore have: S + 4 + S + 6 = 4S
2S + 10 = 4S
2S - 4S = - 10
- 2S = - 10
S, or smallest odd integer =  , or 
The 3 other odd integers are:  ,  , and
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Check
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9 + 11 = 4(5)
20 = 20 (TRUE)
Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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