Question 415702
x+y+z=100 (1)
y=2x+10 (2) 2nd number (y) is (=) 10 more (+) than twice (x) the 1st
z=x+y-10 (3) 3rd number (z) is 10 less (-) than sum of first two (x+y)
Solve for x+y in (3) -> x+y=z+10 
Insert into (1) -> (z+10)+z=100 -> 2z+10=100 -> z=45
From (1) x+y=100-z -> x+y=55 -> y=55-x
Subtitute into (2) -> 55-x=2x+10 -> 3x=45 -> x=15
Since x+y+z=100 -> 15+y+45=100 -> y=40
Ans: x=15,y=40,z=45