The equations to be solved are x^2 + y^2 = 29 and x+ y = 7

x + y = 7

=> y = 7 - x

Substitute in x^2 + y^2 = 29

=> x^2 + (7 - x)^2 = 29

=> x^2 + 49 + x^2 - 14x = 29

=> 2x^2 -14x + 20 = 0

=> 2x^2 - 10x - 4x + 20 = 0

=> 2x (x - 5) - 4(x - 5) = 0

=> (2x - 4)(x - 5) = 0

=> x = 2 and x = 5

=> y = 5 and y= 2

**The solution to the equation is (2, 5) and (5,2)**

The system of equations x^2 + y^2 = 29 and x+ y = 7 has to be solved for x and y.

In the equation x^2 + y^2 = 29, eliminate one of the variables using the other equation. From x + y = 7 we can write x = 7 - x

This gives:

(7 - y)^2 + y^2 = 29

49 + y^2 - 14y + y^2 = 29

2y^2 - 14y + 20 = 0

y^2 - 7y + 10 = 0

y^2 - 5y - 2y + 10 = 0

y(y - 5) - 2(y - 5) = 0

(y - 2)(y - 5) = 0

y = 2 and y = 5

For y = 2, x = 7-y = 5

and for y = 5, x = 7 - y = 2

The solution of the system of equations is x = 2, y = 5 and x = 5, y = 2

This is a symmetric system and we'll solve it using the sum and the product.

We'll note x + y = S and x*y = P

x^2 + y^2 = (x+y)^2 -2xy

x^2 + y^2 = S^2 - 2P

We'll re-write the system in S and P:

S^2 - 2P = 29 (1)

S = 7 (2)

We'll substitute (2) in (1):

49 - 2P = 29

We'll subtract 49 both sides:

-2P = 29 - 49

-2P = -20

We'll divide by -2:

P = 10

We'll substitute P in (1):

S^2 - 20 = 29

We'll add 29 both sides:

S^2 = 49

S1 = 7

S2 = -7

We'll compute x and y:

For S1 = 7 and P = 10

x + y = 7

xy = 10

We'll write the quadratic when we know the sum and the product:

x^2 - 7x + 10 = 0

x1 = [7 + sqrt(49-40)]/2

x1 = (7+3)/2

x1 = 5

x2 = [7 - sqrt(49-40)]/2

x2 = (7-3)/2

x2 = 2

**The solutions of the symmetric system are: {(5 ; 2) ; (2 ; 5)}.**