Question 101503
Question:

 Three consecutive integers are such that the sum of the first and third is 11 less than three times the second. Find the numbers.

Answer:


Here you have three consecutive number...
 can find out the th

So you can take them as,   a, a+1 and a+2


Sum of the first and the third is a + ( a+2)


Three times the second can be written as 3(a+1)


It is given that, the sum of the first and third is 11 less than three times the second

So... a + ( a+2) = 3(a+1)-11


==> a + a + 2= 3a + 3 -11


==> 2a + 2 = 3a + 3 -11


Subtract 3a from both sides..


==> 2a +2 - 3a = 3a -8 - 3a
 

==> -1a + 2 = -8



==> -1a + 2 - 2 = -8 - 2

 
==> -1a  = -10

Divide both sides by -1


==> {{{ -1a/-1 = -10/-1}}}


==> {{{ a = 10 }}}



a = 10


a+1 = 10 +1 = 11


a + 2 = 10 + 2 = 12


So 10,11 and 12 are the three consecutive numbers.   



Hope you found the explanation useful.



Regards.


Praseena.