Question 898326
THE ALGEBRA WAY:
{{{n}}}= the smaller of the two consecutive integers
{{{n+1}}}= the largerer of the two consecutive integers
The problem says that
{{{n/3+(n+1)/4=9}}}
Multiplying both sides of the equalsign times {{{3*4=12}}} we get
{{{12(n/3+(n+1)/4)=12*9}}}
{{{12*n/3+12(n+1)/4=108}}}
{{{4n+3(n+1)=108}}}
{{{4n+3n+3=108}}}
{{{7n+3=108}}}
{{{7n=108-3}}}
{{{7n=105}}}
{{{n=105/7}}}
{{{highlight(n=15)}}}
The two consecutive integers are {{{highlight(15)}}} and {{{highlight(16)}}} .
 
ANOTHER WAY (Guess and check):
One third of the first integer must be an integer (a whole number),
and one fourth of the second integer must be an integer too.
Otherwise, one or both would be fractions that would add up to a fraction.
The first integer could be 0, 3, 6, 9, 12, 15, 18, 21, 24, ....,
and one third of that would be 0, 1, 2, 3, 4, 5, 6, 7, 8, ....
If we add 1 to each of those choices, the second integer would be 1, 4, 7, 13, 16, 19, 22, 25, ...., but not all of those numbers divide by 4 evenly.
Only 4, and 16 do.
If the second integer were 4 and the first one were 3,
one third of the first integer and one fourth of the second integer would be
{{{3/3+4/4=1+1=2}}} , so 3 and 4 is not the answer.
If the second integer were 16 and the first one were 15,
one third of the first integer and one fourth of the second integer would be
{{{15/3+16/4=5+4=9}}} , so {{{highlight(15)}}} and {{{highlight(16)}}} is the answer.