Question 122332
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Let the first of three consecutive integers be {{{n}}}.

Then the second is {{{n+1}}} and the third is {{{n+2}}}.


The sum of the first and third is {{{n+(n+2)}}} or {{{2n+2}}}.
5 times that is {{{5(2n+2)=10n+10}}}

This quantity is (meaning equals) 14 greater than 8 times the second.
8 times the second is {{{8(n+1)=8n+8}}}, and 14 more than that is
{{{8n+8+14=8n+22}}}, so we can write:


{{{10n+10=8n+22}}}


Add {{{-8n}}} to both sides
{{{10n-8n+10=8n-8n+22}}}
{{{2n+10=22}}}


Add {{{-10}}} to both sides
{{{2n+10-10=22-10}}}
{{{2n=12}}}


Multiply by {{{1/2}}}
{{{(1/2)2n=(1/2)12}}}
{{{n=6}}}


Therefore the first number is {{{6}}}, the second is {{{7}}} and the third is {{{8}}}.

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<i>Check the answer</i>


6 plus 8 = 14.   5 times 14 = 70
8 times 7 = 56.  56 plus 14 = 70

Answer checks 






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