To solve the system of equations, we can use the method of substitution.

From equation (1), we have xy = 240.

We can rewrite equation (2) as (x+3)(y+1) = 315. Expanding this equation, we get xy + x + 3y + 3 = 315.

Substituting xy = 240 into this equation, we have 240 + x + 3y + 3 = 315.

Simplifying this equation, we get x + 3y = 72.

Now we have a system of two equations:
xy = 240 (equation 1)
x + 3y = 72 (equation 2)

We can solve equation 2 for x:
x = 72 - 3y

Substituting this expression for x into equation 1, we have (72 - 3y)y = 240.

Expanding this equation, we get 72y - 3y^2 = 240.

Rearranging this equation, we get 3y^2 - 72y + 240 = 0.

Dividing through by 3, we get y^2 - 24y + 80 = 0.

Factoring this quadratic equation, we have (y - 4)(y - 20) = 0.

Setting each factor equal to zero, we have y - 4 = 0 or y - 20 = 0.

Solving these equations, we find y = 4 or y = 20.

Substituting these values of y back into equation 2, we can solve for x:
For y = 4, we have x + 3(4) = 72, which gives x = 60.
For y = 20, we have x + 3(20) = 72, which gives x = 12.

Therefore, the solutions to the system of equations are (x, y) = (60, 4) and (x, y) = (12, 20).