# If x^2 + y^2 = 29 and x+ y = 7 Then find x and y.

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Given the equations:

x^2 + y^2 = 29..............(1)

x+ y = 7.......................(2)

We have a system of two equations and two variables. Then, we can use the substitution or the elimination method to solve.

Let us use the substitution method to solve.

We will re-write equation (2).

x+ y = 7

==> y = 7 - x

Now we will substitute in (1).

x^2 + y^2 = 29

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

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

==> 2x^2 - 14x + 49 - 29 = 0

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

Now we will divide by 2:

==> x^2 - 7x + 10 = 0

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

==> x1 = 2 ==> y1= 7-2 = 5

==> x2= 5 ==> y2= 7-5 = 2

**Then the answer is the pairs:**

**( 2, 5) OR ( 5, 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)}.**

If x^2 + y^2 = 29 and x+ y = 7. To find x and y.

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

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

2xy = 7^2-29 = 49-29 = 20.

(x-y)^2 = (x+y)^2-4xy

(x-y)^2 = 7^2-2(20) = 49-40 = 9.

Therefore x-y = sqrt9 = 3.

Thus we have x+y = 7....(1)

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

(1)+(2) gives: 2x= 7+3 = 10. So x= 10/2 = 5.

(1)-(2) gives: 7-3 = 4. So 2y = 4. Or y = 4/2 = 2.

Therefore x= 5 and y = 2.