Question

Alguém me ajudar nessas 2 questões Pfv

Answer 1

Resposta:

1) correta é a

2) correta é e

Explicação passo-a-passo:

1)

$\frac{{\left \{ x+y+z=6} \atop {2x+y-z=1} \right.\\}{3x+2y=7}$

$\frac{{\left \{ 2x+y-z=1} \atop {3x-y+z=4} \right.\\}{5x=5}$

$x=1$

$3x+2y=7\\3(1)+2y=7\\3+2y=7\\2y=7-3\\2y=4\\y=\frac{4}{2} \\y=2$

x+y+z=6

1+2+z=6

3+z=6

z=6-3

z=3

S={(x,y,z)}={(1,2,3)}

2)

$\left[\begin{array}{ccc}3x-2y+z=-6\\4y-2z=1\\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x-2y+z=-6\\4y-2(2)=1\\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x-2y+z=-6\\4y-4=1\\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x-2y+z=-6\\4y=1+4\\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x-2y+z=-6\\4y=5\\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x-2y+z=-6\\y=\frac{5}{4} \\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x-2(\frac{5}{4}) +2=-6\\y=\frac{5}{4} \\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x-2*\frac{5}{4} =-6-2\\y=\frac{5}{4} \\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x-\frac{5}{2} =-8\\y=\frac{5}{4} \\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x=-8+\frac{5}{2}\\y=\frac{5}{4} \\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x=-\frac{16}{2} +\frac{5}{2}\\y=\frac{5}{4} \\z=2\end{array}\right]$

$\left[\begin{array}{ccc}3x=-\frac{11}{2}\\y=\frac{5}{4} \\z=2\end{array}\right]$

$\left[\begin{array}{ccc}x=-\frac{11}{2}*\frac{1}{3} \\y=\frac{5}{4} \\z=2\end{array}\right]$

$\left[\begin{array}{ccc}x=-\frac{11}{6} \\y=\frac{5}{4} \\z=2\end{array}\right]$

$x=-\frac{11}{6}$≅-1,8333...≅-2

$\\y=\frac{5}{4}$=1,25≅1

$z=2$

S={(x,y,z)}={(-2,1,2)}

mais sobre o assunto:https: //www.youtube.com/watch?v=hP59QyuhTbc