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

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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)

 

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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)

 

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