The sum of 3 numbers is 78,
Let the 1st number = x
Let the 2nd number = y
Let the 3rd number = z
x + y + z = 78
the third 4 more than the second,
z = y + 4
the first is two less than twice the second
x = 2y - 2
Substitute (2y - 2) for x, and (y + 4) for z in x + y + z = 78
(2y - 2) + y + (y + 4) = 78
Solve that for y, then substitute what you get for y in
x = 2y - 2 to find x, and in z = y + 4 to find z.
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