3x^2 + 2y^2 - 54y - 143 = 0............(1)

x-3y -3 = 0............(2)

Let us rewrite equation (2) :

x - 3y - 3 = 0

==> x= 3y + 3= 3(y+1)

==> x= 3(y+1)

Now we will substitute x value in (1):

3x^2 + 2y^2 - 54y - 143 = 0

3(3(y+1))^2 + 2y^2 - 54y - 143 = 0

3(9(y^2 + 2y+1) + 2y^2 - 54y - 143 = 0

27y^2 + 54y + 27 + 2y^2 - 54y - 143 = 0

Now combine like terms:

29y^2 - 116 = 0

==> 29y^2 = 116

Now divide by 29:

==> y^2 = 4

**==> y= +- 2**

**==> x= 3(y+1)**

**==> x1= 3(2+1) = 9**

**==> x2= 3(-2+1) = -3**

**==> the answer is:**

**(9, 2) and (-3, -2) **

We have to solve for x and y using 3x^2 + 2y^2 - 54y - 143 = 0 and x-3y -3 = 0

Let us write x-3y -3 = 0 as x = 3y + 3

Therefore 3x^2 + 2y^2 - 54y - 143 = 0

=> 3(3y + 3)^2 + 2y^2 - 54y - 143 =0

=> 27y^2 + 54y +27 + 2y^2 - 54y - 143 =0

=> 29y^2 - 116 = 0

=> y^2 = 116/29 = 4

Therefore y can be +2 or -2

As x = 3y + 3, for y = +2, x= 9 and for y=-2, x=-3

**So the values of x and y are (-3, -2) and (9, 2).**