Question

Resolva o sistema
2x+y^2=4
2x-y=2​

Answer 1

Olá!

2x- y= 2

y= 2x-2

2x+ y² =4

2x+ (2x-2)² = 4

2x + 4x² -8x +4 = 4

4x² -6x= 4-4

4x² -6x= 0

2x(2x - 3) = 0

2x= 0

x= 0

2x - 3 = 0

2x= 3

x= 3/2

y= 2x-2

y= 2 * 0 -2

y= -2

y= 2x-2

y= 2*(3/2) -2

y= 1

Resposta:

Solução: [(3/2), 1]

              [0, -2)

Bons estudos!

T.H.S

Answer 2

2x+y²=4

2x-y=2--->2x=2+y

2x+y²=4

2+y+y²=4

y²+y-2=0

∆=1-4.1.-2

∆=9

y=-1±√9/2

y=-1±3/2

y1=-1+3/2

y1=2/2=1

y2=-1-3/2

y2=-4/2=-2

2x=2+y1

2x=2+1

x1=2/3

2x=2+y2

2x=2-2

x2=0

(x1,y1)=(3/2,1)

(y1+y2)=(0,-2)

Bons estudos :)