Solve the simultaneous equations x^3+y^3=-7 and x^2+y^2=5?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The equations to be solved are:

x^3 + y^3 = -7 ...(1)

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

x^3 + y^3 = -7

=> x^3 + y^3 = -8 + 1 = 1 - 8

=> x^3 + y^3 = (-2)^3 + 1^3  = 1^3 + (-2)^3...(3)

x^2 + y^2 = 5

=> x^2 + y^2 = 4 + 1 = 1 + 4

=> x^2 + y^2 = (-2)^2 + 1^2 = 1^2 + (-2)^2 ...(4)

From (3) and (4), we can see that substituting x with -2 and y with 1 and x with 1 and y with -2 solves both the equations.

The solution of the simultaneous equations are (-2 , 1) and (1, -2)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll apply the formula of the sum of the cubes:

x^3 + y^3 = (x+y)(x^2 - xy + y^2)

We'll replace the sum x+y by S and the product x*y by P.

The system will become:

S(5 - P) = -7 (1)

S^2 - 2P = 5

S^2 = 5 + 2P (2)

We'll raise to square (1):

S^2*(5-P)^2 = 49

(5 + 2P)(5-P)^2 = 49

We'll raise to square the binomial:

(5 + 2P)(25-10P + P^2) = 49

125 - 50P + 5P^2 + 50P - 20P^2 + 2P^3 = 49

We'll eliminate like terms:

-15P^2 + 2P^3 + 76 = 0

We notice that one root is P = -2

-60 - 16 + 76 = 0

We'll re-write 2P^3 - 15P^2 + 76 = (P+2)(2P^2 - 19P + 38)

We'll cancel 2P^3 - 15P^2 + 76 = 0

(P+2)(2P^2 - 19P + 38) = 0

2P^2 - 19P + 38 = 0

P1 = [19+sqrt(361 - 304)]/4

P1 = (19+sqrt 57)/4

P2 = (19 - sqrt 57)/4

Since S(5 - P) = -7 then we'll have:

P = -2 => S(5+2)=-7 => S = -1

P = (19+sqrt 57)/4 => S = (1+sqrt57)/2

P = (19-sqrt 57)/4 => S = (1-sqrt57)/2

Now, we'll solve the systems:

1) x + y = -1 and xy = -2

x^2 + x - 2 = 0

x = -2 and y = 1 or x = 1 and y = -2

2) x + y = (1+sqrt57)/2 and x*y = (19+sqrt 57)/4

x^2 - (1+sqrt57)*x/2 + (19+sqrt 57)/4 = 0

delta = (1+sqrt57)^2 - 4*19-4*sqrt 57

delta = 1 + 2sqrt 57 + 57 - 76 - 4*sqrt57

delta = -18 - 2sqrt 57 < 0

Since the discriminant is negative, this system has no solutions.

Since the system is symmetrical, the solutions of the system of equations are represented by the pairs: (1 , -2) and (-2 , 1).

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