Question

Resolva os sistema(U=R)

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Answer 1

Resposta:

a)

Isola x em:

x + y = 7

x = 7 - y

Substituir x

2x² - y² = 46

2(7 - y)² - y² = 46

2(49 - 14y +y²) - y² = 46

98 - 28y + 2y² - y² = 46

y² - 28y + 98 - 46 = 0

y² - 28y + 52 = 0

a = 1

b = -28

c = 52

Δ = b² -4ac

Δ = (-28)²  - 4(1)(52)

Δ = 784 - 208

Δ = 576

$y=\dfrac{-b\pm\sqrt{\Delta} }{2a}=\dfrac{28\pm24}{2}\\\\ \\ y_1=\dfrac{28+24}{2}=\dfrac{52}{2}=26\\ \\ \\ y"=\dfrac{28-24}{2}=\dfrac{4}{2}=2$

Como temos dois valores para y, também teremos dois valores para x

se y = y₁ = 26

Sendo:

x = 7 - y'

x'= 7 - 26

x"= -19

x" = 7 -2

x"= 5

S = { ( -19, 26);(5 , 2 )}

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b)

Isola x em:

x - y = 2

x = 2 + y

Substituir x = 2 + y em:

x² + y² = 34

( 2 + y)² + y² = 34

4 + 4y + y² -34=0

2y² + 4y - 30 = 0   → :2

y² + 2y -15 = 0

a = 1

b = 2

c = -15

Δ = b² - 4ac

Δ = 2² -4.1.(-15)

Δ = 4 + 60

Δ = 64

$y=\dfrac{-2\pm8}{2}\\ \\ \\ y'=\dfrac{-2+8}{2}=\dfrac{6}{2}=3\\ \\ \\ y"=\dfrac{-2-8}{2}=-\dfrac{10}{2}=-5$

Se x = 2 + y

x' = 2 + 3

x' = 5

x"= 2 - 5

x" = -3

S = { ( 5 , 3 ); (-3 , -5 )}

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c)

$isola ~~x~~em~~x.y=2\\ \\ x=\dfrac{2}{y}$

Substituir x em 2x - 3y = 1

$2.(\dfrac{2}{y})-3y=1\\ \\ \\ 4-3y^2=y\\ \\ -3y^2-y+4=0~~~\times(-1)\\ \\ 3y^2+y-4=0$

a = 3

b = 1

c = -4

Δ = b² - 4ac

Δ = 1² - 4.(3).(-4)

a = 3

b = 1

c = -4

Δ = b² - 4ac

Δ = 1² - 4.(3).(-4)

Δ = 1 + 48

Δ = 49

 $y=\dfrac{-1\pm7}{6}\\ \\ \\ y'=\dfrac{-1+7}{6}=\dfrac{6}{6}=1\\ \\ \\ y"=\dfrac{-1-7}{6}=-\dfrac{8}{6}=-\dfrac{4}{3}$

$Se~~x =\dfrac{2}{y}\\ \\ \\ x'=\dfrac{2}{1}=2\\ \\ \\ x"=\dfrac{2}{-\dfrac{4}{3}}=2\times(-\dfrac{3}{4})=-\dfrac{6}{4}=-\dfrac{3}{2}$

$S=\{(1,2);(-\dfrac{3}{2},-\dfrac{4}{3})\}$