Question 791444: Find three consecutive integers such that five times the third integer equals twice the first integer plus seven more than the second integer?
Found 4 solutions by CubeyThePenguin, MathTherapy, josgarithmetic, greenestamps:
Answer by CubeyThePenguin(3113)   (Show Source):
You can put this solution on YOUR website! consecutive integers: (x-1), x, (x+1)
5(x+1) = 2(x-1) + 7x
5x + 5 = 9x - 2
7 = 4x
The solutions of this equation are not integers.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
Find three consecutive integers such that five times the third integer equals twice the first integer plus seven more than the second integer?
He's WRONG, one AGAIN!! Integers DO exist!!
Answer by josgarithmetic(39617)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Oh, my........
Tutor @CubeyThePenguin: wrong equation, leading to wrong answer.
Tutor @MathTherapy: nothing of use to you the student....
Tutor @josgarithmetic: algebra errors, leading to wrong answer... (which she could have seen if she had bothered to check to see if her answers were right).
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While there are many problems involving three consecutive integers where it is advantageous to call the numbers x-1, x, and x+1, I don't see that being helpful here. So let the numbers be x, x+1, and x+2. Then
"...five times the third integer equals twice the first integer plus seven more than the second integer"
5 times the third integer: 5(x+2)
twice the first integer plus 7 more than the second integer: 2x+((x+1)+7)
ANSWER: The three consecutive integers are -1, 0, and 1.
CHECK:
5 times the third integer: 5(1) = 5
twice the first integer plus 7 more than the second integer: 2(-1)+(0+7) = -2+7 = 5
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