# Solve the system : x + y = 5 x^2 + y^2 = 13

*print*Print*list*Cite

### 4 Answers

We have to solve x+y = 5 and x^2 + y^2 = 13 for x and y .

First we take x+y = 5

square both the sides

=> (x + y)^2 = 5^2

=> x^2 + y^2 + 2xy = 25

Now x^2 + y^2 = 13

=> 13 + 2xy = 25

=> 2xy = 12

=> xy = 6

=> x = 6/y

replace in x + y = 5

=> 6/y + y = 5

=> 6 + y^2 = 5y

=> y^2 - 5y + 6 = 0

=> y^2 - 3y - 2y + 6 =0

=> y (y - 3 ) - 2( y -3) = 0

=> (y - 2)(y-3) =0

So y = 2 or 3

As x = 6 /y , x = 3 or 2

**Therefore the result is (3 , 2) and ( 2,3)**

x+ y= 5...............(1)

x^2 + y^2 = 13................(2)

First we will rewrite (1) as a function of x:

==> y= 5 - x

Now we will use the substitution method to solve:

==> x^2 + y^2 = 13

==> x^2 + ( 5-x)^2 = 13

==> x^2 + 25 - 10x + x^2 = 13

==> 2x^2 - 10x + 25 -13 = 0

==> 2x^2 - 10x + 12 = 0

Now we will divide by 2:

==> x^2 - 5x + 6 = 0

==> ( x -3) ( x-2) = 0

==> x1= 3 ==> y1= 2

==> x2= 2 ==> y2= 3

Then the answer is :

(** 2,3) OR ( 3,2)**

To solve the sytem of equations:

x+y = 5 ..........(1) and

x^2+y^2 = 13......(2).

We we know 2xy = (x+y)^2 - (x^2+y^2).

Therefore 2xy = 5^2-13 = 25-13 = 12.

Therefore (x-y)^2 = x^2+y^2-2xy = 13 -12 = 1.

Therefore x-y = sqrt {(x-y)^2 } = sqrt1 or -sqrt1.

x-y = 1. Or x-y = -1.

Or x-y = 1....(3).

x+5 = 5........(1).

(1)+(3):

2x = 1+5= 6.

x = 6/2 = 3.

(1)-(3):

2y = 5-1 = 4.

y = 4/2 = 2.

Therefore the solution of the system of equations: x= 3, y = 2, O

Or

By using the equations x-y = -sqrt1 = -1, x+y = 5, we get x = 2 and y = 3.

The set of equations x + y = 5 and x^2 + y^2 = 13 has to be solved.

From x + y = 5, we get x = 5 - y

Substitute this in x^2 + y^2 = 13

(5 - y)^2 + y^2 = 13

25 - 10y + y^2 + y^2 = 13

2y^2 - 10y + 12 = 0

2y^2 - 6y - 4y + 12 = 0

2y(y - 3) - 4(y - 3) = 0

(2y - 4)(y - 3) = 0

2y - 4 = 0

y = 4/2 = 2

y - 3 = 0

y = 3

As x = 5 - y

For y = 2, x = 3

For y = 3, x = 2

The solution of the given set of equations is (2, 3) and (3,2)