Question

x²+y²+z²=29
-x+y+2z=9
x+y+z=9
quais são as soluções de x,y e z​

Answer 1

$\displaystyle \sf (1) \ x^2+y^2+z^2=29 \\\\ (2)\ -x+y+2z = 9\\\\ (3) \ x+y+z=9 \\\\\\ \underline{\text{Note que as equa{\c c\~oes}(2) e (3) s{\~a}o iguais a 9, logo podemos igualar }}: \\\\ -x+y+2z = x+y+z \\\\ \boxed{\sf 2x = z } \\\\\\ \underline{\text{Somando as equa{\c c\~oes} (2) e (3)}}: \\\\ -x+y+2z+x+y+z = 9+9 \\\\ 2y+3\underbrace{\sf z}_{2x} = 18 \\\\ 2y+3(2x) = 18 \\\\ 2y +6x=18 \\\\ \boxed{\sf y = 18-3x}$

Agora vamos substituir os valores de z e de y em função x na equação (1) :

$\displaystyle \sf x^2+y^2+z^2 = 29 \\\\ x^2+(9-3x)^2+(2x)^2=29 \\\\ x^2+81-54x+9x^2+4x^2 -29= 0 \\\\ 14x^2-54x+52=0\ \ (\div 2) \\\\ 7x^2-27x+26 = 0 \\\\ 7x^2-14x-13x+26 = 0\\\\ 7x(x-2)-13(x-2) = 0\\\\ (x-2)(7x-13)= 0 \\\\ 7x-13 = 0 \to \boxed{\sf x = \frac{13}{7}} \\\\\\ x-2 = 0 \to \boxed{\sf x =2}\\\\ \underline{\text{Usando x = 2 , temos }}:$

$\sf z =2x \to z=2.2 \to \boxed{\sf z = 4}\\\\ y=9-3x\to y=9-3.2 \to \boxed{\sf y=3}$

Portanto :

$\displaystyle \sf \boxed{\sf \begin{array}{I}\displaystyle \\ \boxed{\sf x = 2} \\\\ \displaystyle \boxed{\sf y = 3} \\\\ \displaystyle \boxed{\sf z = 4}\\ \ \end{array} }\checkmark$