{(3/2)^x-y -(2/3 ) ^x-y=65/36 {cycle-x+y=118

Solve the system: (3/2)^(x−y) − (2/3)^(x−y) = 65/36 x + y = 118

Let d = x − y. Then (3/2)^d − (2/3)^d = 65/36. Let t = (3/2)^d > 0, so (2/3)^d = t^(-1). The equation becomes: t − t^(-1) = 65/36 ⇒ t^2 − (65/36)t − 1 = 0 ⇒ 36t^2 − 65t − 36 = 0

Discriminant: Δ = 65^2 + 4·36·36 = 9409 = 97^2 t = [65 ± 97]/72

Positive root: t = (65 + 97)/72 = 162/72 = 9/4. (The other root t = -4/9 is invalid since t > 0.)

Thus (3/2)^d = 9/4 = (3/2)^2, so d = x − y = 2.

Now solve: x + y = 118 x − y = 2

Adding: 2x = 120 ⇒ x = 60 Then y = 118 − x = 58

Answer: x = 60, y = 58.