Question

Encontre a solucao do sistema abaixo

${2x}^{2} - {y}^{2} = 7$
$x = y - 1$

Answer 1

2x^2 - y^2 = 7

x = y - 1

2.(y - 1)^2 - y^2 = 7

2.(y^2 - 2.1.y + 1) - y^2 = 7

2.(y^2 - 2y + 1) - y^2 = 7

2y^2 - 4y + 2 - y^2 = 7

y^2 - 4y + 2 - 7 = 0

y^2 - 4y - 5 = 0

a = 1; b = - 4; c = - 5

∆ = b^2-4ac

∆ = (-4)^2 - 4.1.(-5)

∆ = 16 + 20

∆ = 36

y = [ - b +/- √∆]/2a

y = [ -(-4) +/- √36]/2.1

y = [ 4 +/- 6]/2

y' = [4 + 6]/2 = 10/2 = 5

y" = [4 - 6]/2 = -2/2 = - 1

y = 5

x = y - 1

x = 5 - 1

x = 4

y = - 1

x = y - 1

x = - 1 - 1

x = - 2

R.:

x = - 2; y = - 1

x = 4; y = 5