System solvingSolve the system x+y=5 x^2+y^2=13

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve

x + y = 5 ...(1)

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

From (1) x = 5 - y

Substituting in (2)

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

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

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

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

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

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

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

=> y = 2 and y = 3

x = 3 and 2

The solutions of the equations are (2, 3) and (3,2)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

We'll note x+y = S and xy = 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 = 5

S^2 - 2P = 13

We'll susbtitute S in the second equation:

25 - 2P = 13

-2P = 13 - 25

-2P = -12

P = 6

Now, we'll form the quadratic knowing the sum and the product:

x^2 - 5x + 6 = 0

x1 = [5+sqrt(25 - 24)]/2

x1 = (5+1)/2

x1 = 3

x2 = (5-1)/2

x2 = 2

So, the solutions of the symmetric system are: {2 ; 3} or {3 ; 2}.

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