Question 259482
'Find four consecutive integers 
x, (x+1), (x+2), (x+3
:
such that the sum of the two greatest is 17 less than twice the sum of the two smallest.'

 x+2 + x+3 = 2(x + x+1) - 17
:
2x + 5 = 2(2x + 1) - 17
;
2x + 5 = 4x + 2 - 17
:
2x + 5 = 4x - 15
:
5 + 15 = 4x - 2x
:
20 = 2x
x = {{{20/2}}}
x = 10, 11, 12, 13 are the 4 consecutive numbers
;
:
Check solution:
13 + 12 = 2(11+10) - 17
25 = 2(21) - 17
25 = 42 - 17