Given,
a + a = 10
i.e.
2a = 10,
so a = 5.......................................(i)
a + @ = 15,
@ = 15 - a, but a = 5,
so @ = 10....................................(ii)
a + e = 25,
e = 25 - a, but a = 5,
so, e = 20.....................................(iii)
adding the three equations above, we have
a + @ + e = 5 + 10 + 20
which gives,
a + @ + e = 35