Question 885747
Let the ages of the sons be x and y
x-Older son
y-Younger son


Let the fathers age be z


10 years ago their ages would have been
(x-10), (y-10) and (z-10)


10 years ago,The ages of two sons equals 1/3 of their fathers 
(x-10)+(y-10)=1/3*(z-10)


Simplify and get...3x+3y-z=50--------------------(1)


Since one son is 2 years older, x=y+2----------------(2)


Currently the sum of the children's ages is the Father's less 14 meaning
x+y = z-14-------------------------------------------------(3)


if you substitute eqn 2 in 3 you get
z=2y+16

Now go back to equation 1 and make sure its all in terms of y

3(y+2)+3y-(2y+16) = 50
4y=60
y=15


therefore x=2+15 =17


z=2*15+16 = 46



Their present ages are 15,17,46