Question

Se o termo (x0, y0, z0) é a solução do sistema abaixo, então determine o valor de 3x0+5y0+4z0.

3x+z= -5
x+y+z= -2
2y-Z= -3

Answer 1

$\begin{cases} \mathrm{3x+z=-5\ \to\ x=\dfrac{-5-z}{3}}\\ \mathrm{x+y+z=-2}\\ \mathrm{2y-z=-3\ \to\ y=\dfrac{z-3}{2}}\\ \end{cases}\\\\\\ \mathrm{x+y+z=-2\ \to\ \dfrac{-5-z}{3}+\dfrac{z-3}{2}+z=-2}\\\\ \mathrm{\dfrac{2(-5-z)+3(z-3)+6z}{6}=-2}\\\\ \mathrm{-10-2z+3z-9+6z=-12\ \to\ 7z-19=-12}\\\\ \mathrm{7z=-12+19\ \to\ 7z=7\ \to\ \boxed{\mathbf{z=1}}}\\\\ \mathrm{x=\dfrac{-5-z}{3}=\dfrac{-5-1}{3}=\dfrac{-6}{3}\ \to\ \boxed{\mathbf{x=-2}}}\\\\ \mathrm{y=\dfrac{z-3}{2}=\dfrac{1-3}{2}=\dfrac{-2}{2}\ \to\ \boxed{\mathbf{y=-1}}}\\\\\\ \mathrm{(x_0,y_0,z_0)=\boxed{\mathbf{(-2,-1,1)}}}\\\\ \mathrm{3x_0+5y_0+4z_0=3.(-2)+5.(-1)+4.1=}\\\\ \mathrm{=-6-5+4=-11+4=\boxed{\boxed{\mathbf{-7}}}}$