Question 1093360: How do you find 3 consecutive numbers such that 3 times the 2nd number is 93 more than the largest number? Found 4 solutions by josgarithmetic, MathTherapy, greenestamps, richwmiller:Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website! -----------------------------------------------------------------------------
How do you find 3 consecutive numbers such that 3 times the 2nd number is 93 more than the largest number?
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More than one way to do this, but give a variable to one of the consecutive numbers.
n, first number
n+1 and n+2, the next two numbers
Translate the description exactly as it was written:
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BETTER WAY:
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Let n be the asked-for "largest number".
The numbers are therefore n-2, n-1, and n. -----"largest number"
**************************************************** Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
How do you find 3 consecutive numbers such that 3 times the 2nd number is 93 more than the largest number?
The NON-CONFUSING method:
Let the 1st number be F
Then others are F + 1, and F + 2
We then get: 3(F + 1) = F + 2 + 93
3F + 3 = F + 95
3F - F = 95 - 3
2F = 92
F, or 1st number =
Others:
That's all....nothing CONFUSING or COMPLEX!
"3 times a number is 93 more than the next number"
Since we ignored the "first number" mentioned in the problem, our answer of 47 is the "second" number; so the three numbers are 46, 47, and 48. Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website! Nothing confusing nor complex about josgarithmetic's solution. You just don't like him. Get over yourself, MathTherapy. I do believe you need another type of therapy.