The ages of three brothers add up to 38 years. The oldest brother is 5 years older than the youngest brother. Five years ago, the oldest was twice the age of the youngest. Write and solve a system of three equations in three variables to determine the ages of the brothers.

Let y be the age of the youngest.

Let the age of the oldest be x.

x = y + 5 ... (1)

2 (y - 5) = x - 5 ... (2)

Plugging the expression for x in (1) into (2), we get:

2 (y - 5) = y + 5 - 5, which gives:

2y - 10 = y

y = 10

Plugging the value y = 10 into (1), we get:

x = 15.

Let the age of the 'middle' brother be m.

x + y + m = 38 ... (3)

Plugging the values for x and y into (3), gives:

10 + 15 + m = 38

m = 38 - (10 + 15) = 13

The brothers ages are 10, 13 and 15.