Question

resolva o sistema de equações.​

Answer 1

Explicação passo-a-passo:

X² + y² = 20

X + y = 6 → X = 6 - y

Por meio da substituição;

(6-y)² + y² = 20

36 -12y + y² +y² = 20

2Y² -12y +16 = 0

Y² - 6Y + 8 = 0

Y = -b±√B²-4ac/2a

Y= 6±√36-32/2

Y= 6±√4/2 = 6±2/2

Y = 4 ou Y = 2

Voltando no X = 6-y

X= 6-2

ou X = 6-4

Quando Y = 2, X = 4

Quando Y=4, X = 2

B) X² + 2Y² = 18

X- Y = -3 → X = Y-3

substituindo X = Y-3 no sistema:

(Y-3)²+2Y² = 18

Y²-6Y+9 + 2Y² = 18

3Y²-6Y-9 = 0

Y²-2Y-3 = 0

Y= -(-2)±√4+12/2

Y=2±√16/2

Y=2±4/2

Y= 3 ou Y = -1

Voltando no X= Y-3,

X= (3)-3 = 0

X= (-1)-3 = -4

Quando o Y= 3, X= 0

Quando o Y= -1, X= -4

Answer 2

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$\tt a)\begin{cases}\sf x^2+y^2=20\\\sf x+y=6\end{cases}\\\begin{cases}\sf x^2+y^2=20\\\sf y=6-x\end{cases}\\\sf x^2+(6-x)^2=20\\\sf x^2+36-12x+x^2-20=0\\\sf 2x^2-12x+16=0\div2\\\sf x^2-6x+8=0\\\sf\Delta=b^2-4ac\\\sf \Delta=(-6)^2-4\cdot1\cdot8\\\sf\Delta=36-32\\\sf\Delta=4\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-(-6)\pm\sqrt{4}}{2\cdot1}\\\sf x=\dfrac{6\pm2}{2}\begin{cases}\sf x_1=\dfrac{6+2}{2}=\dfrac{8}{2}=4\\\sf x_2=\dfrac{6-2}{2}=\dfrac{4}{2}=2\end{cases}$

$\sf y=6-x\Bigg|_{x=4}\implies y=6-4=2\\\sf y=6-x\Bigg|_{x=2}\implies y=6-2=4\\\huge\boxed{\boxed{\boxed{\boxed{\sf S=\{(2,4),(4,2)\}}}}}$

$\tt b)\begin{cases}\sf x^2+2y^2=18\\\sf x-y=-3\end{cases}\\\begin{cases}\sf x^2+2y^2=18\\\sf y=x+3\end{cases}\\\sf x^2+2\cdot(x+3)^2=18\\\sf x^2+2\cdot( x^2+6x+9)=18\\\sf x^2+2x^2+12x+\diagup\!\!\!\!\!\!18=\diagup\!\!\!\!\!\!18\\\sf 3x^2+12x=0\\\sf 3x\cdot(x+4)=0\\\sf 3x=0\implies x=\dfrac{0}{3}=0\\\sf x+4=0\\\sf x=-4$

$\sf y=x+3\Bigg|_{x=0}\implies y=0+3=3\\\sf y=x+3\Bigg|_{x=-4}\implies y=-4+3=-1\\\huge\boxed{\boxed{\boxed{\boxed{\sf s=\{(0,3),(-4,-1)\}}}}}$