a.
{7x^2 + 13y = -39
{5x^2 - 11y = 33

From the first equation, we can solve for y:
13y = -39 - 7x^2
y = (-39 - 7x^2) / 13

Substitute y into the second equation:
5x^2 - 11((-39 - 7x^2) / 13) = 33
5x^2 + (429 + 77x^2) / 13 = 33
65x^2 + 429 + 77x^2 = 429
142x^2 = 0
x = 0

Substitute x = 0 into the first equation:
7(0)^2 + 13y = -39
13y = -39
y = -3

Therefore, the solution to the system of equations is x = 0, y = -3.

b.
{2x^2 + y^2 = 10
{x^2 - 2y^2 = 5

From the second equation, we can solve for x:
x^2 = 2y^2 + 5
x = ±√(2y^2 + 5)

Substitute x into the first equation:
2(±√(2y^2 + 5))^2 + y^2 = 10
2(2y^2 + 5) + y^2 = 10
4y^2 + 10 + y^2 = 10
5y^2 = 0
y = 0

Substitute y = 0 into x^2 = 2y^2 + 5:
x^2 = 2(0)^2 + 5
x^2 = 5
x = ±√5

Therefore, the solution to the system of equations is x = ±√5, y = 0.

c.
{ (x+3)^2 - 2y^3 = 6
{ 3(x+2)^2 + 5y^3 = 7

Expand the first equation:
x^2 + 6x + 9 - 2y^3 = 6
x^2 + 6x + 3 - 2y^3 = 0

Expand the second equation:
3(x^2 + 4x + 4) + 5y^3 = 7
3x^2 + 12x + 12 + 5y^3 = 7
3x^2 + 12x + 5y^3 = -5

Substitute x^2 + 6x + 3 - 2y^3 into 3x^2 + 12x + 5y^3 = -5:
3(x^2 + 6x + 3 - 2y^3) + 12x + 5y^3 = -5
3x^2 + 18x + 9 - 6y^3 + 12x + 5y^3 = -5
3x^2 + 30x + 9 - y^3 = -5
3x^2 + 30x + y^3 = -14

There is no clear solution to this system of equations.