Question 1093360
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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(n+1)=(n+2)+93}}}
-
{{{3n+3=n+95}}}
{{{2n=92}}}
{{{n=46}}}
---------------------{{{n+2=highlight(48)}}}




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(n-1)=n+93}}}
{{{3n-3=n+93}}}
{{{2n=96}}}
{{{highlight(n=48)}}}-----"largest number"
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