SOLUTION: 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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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) About Me  (Show Source):
You can put this solution on YOUR website!
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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:
3%28n%2B1%29=%28n%2B2%29%2B93
-
3n%2B3=n%2B95
2n=92
n=46
---------------------n%2B2=highlight%2848%29



BETTER WAY:
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Let n be the asked-for "largest number".
The numbers are therefore n-2, n-1, and n.
3%28n-1%29=n%2B93
3n-3=n%2B93
2n=96
highlight%28n=48%29-----"largest number"
****************************************************

Answer by MathTherapy(10552) About Me  (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 = highlight_green%28matrix%281%2C3%2C+92%2F2%2C+%22=%22%2C+46%29%29

Others: 

That's all....nothing CONFUSING or COMPLEX!

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!

An even easier way to solve the problem is to note that the given information says nothing about the first of the three numbers. So really all you have is

"3 times a number is 93 more than the next number"

3x+=+%28x%2B1%29%2B93
3x+=+x%2B94
2x+=+94
x+=+47

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) About Me  (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.