Question

(7/2)x+(2/5)y-3z+217=0
(4/3)x+(6/7)y-(8/21)z-6=0
5x+3y-(7/4)z+12=0
son ecuaciones lineales

Answer 1

Respuesta:

z=84,y=35,x=6

Explicación paso a paso:

$\begin{bmatrix}\left(\frac{7}{2}\right)x+\left(\frac{2}{5}\right)y-3z+217=0\\ \left(\frac{4}{3}\right)x+\left(\frac{6}{7}\right)y-\left(\frac{8}{21}\right)z-6=0\\ 5x+3y-\left(\frac{7}{4}\right)z+12=0\end{bmatrix}$

Despejaremos x de la primera ecuacion:

$\left(\frac{7}{2}\right)x+\left(\frac{2}{5}\right)y-3z+217=0\\\\\frac{7}{2}x+217=-\left(\frac{2}{5}y-3z\right)\\\\\frac{7}{2}x=-\frac{2}{5}y+3z-217\\\\7x=-\frac{4}{5}y+6z-434\\\\x=\frac{-4y+30z-2170}{35}$

Ahora y:

$\frac{4}{3}\cdot \frac{-4y+30z-2170}{35}+\frac{6}{7}y-\frac{8}{21}z-6=0\\\\\frac{4\left(-4y+30z-2170\right)}{105}+\frac{6}{7}y-6=\frac{8}{21}z\\\\\frac{4\left(-4y+30z-2170\right)}{105}+\frac{6}{7}y=\frac{8}{21}z+6\\\\4\left(-4y+30z-2170\right)+90y=40z+630\\\\74y+120z-8680=40z+630\\\\74y-8680=-80z+630\\\\74y=-80z+9310\\\\y=\frac{5\left(-8z+931\right)}{37}$

Despejar z:

$5\cdot \frac{-4\cdot \frac{5\left(-8z+931\right)}{37}+30z-2170}{35}+3\cdot \frac{5\left(-8z+931\right)}{37}-\frac{7}{4}z+12=0\\\\-\frac{93z}{1036}-\frac{165}{37}+12=0\\\\-93z-4620+12432=0\\\\-93z+7812=0\\\\-93z=-7812\\\\z=84$

Para y sustituir z

$y=\frac{5\left(-8\cdot \:84+931\right)}{37}\\\\y==\frac{1295}{37}\\\\y=35$

Para x sustituir z

$x=\frac{-4\cdot \:35+30\cdot \:84-2170}{35}\\\\x=\frac{210}{35}\\\\x=6$

Ejercicio:

$\begin{bmatrix}x+2y-z=1\\ 2x+y+4z=2\\ 3x+3y+4z=1\end{bmatrix}$

DESPEJAR X:

$x+2y-z=1\\\\x=1-2y+z$

DESPEJAR Y:

$2\left(1-2y+z\right)+y+4z=2\\\\2\left(1-2y+z\right)+y=2-4z\\\\-3y+2z+2=2-4z\\\\-3y+2=2-6z\\\\-3y=-6z\\\\y=2$

DESPEJAR Z:

$3\left(1-2\cdot \:2z+z\right)+3\cdot \:2z+4z=1\\\\3\left(-3z+1\right)+6z+4z=1\\\\3\left(1-3z\right)+10z=\\\\3-9z+10z=1\\\\3+z=1\\\\z=-2$

Para y sustituir z:

$y=2\left(-2\right)\\y=-4$

Para x sustituir y y z:

$x=1-2\left(-4\right)-2\\\\x=7$

z=-2,y=-4,x=7