From the adjacent figure ΔABC, D is a point on BC.

Such that ΔABD ≅ ΔACD
AB = 2x + 3
AC = 3y + 1
BD = x
DC = Y + 1
Δ ABD = Δ ACD
⇒ AB = AC
⇒ 2x + 3 = 3y + 1
⇒ 2x – 3y + 2 = 0 …………….(1)
BD = CD
⇒ x = y + 1
⇒ x – y – 1 = 0 ………………….(2)
⇒ from (1) & (2)

y = 4
from (2) ⇒ x – 4 = 1
⇒ x = 1 + 4 = 5
∴ (x, y) = (5, 4)