Question

como resolver el siguiente ejercicio por metodo de reduccion
x-3y+2z=3
2x+y-5z=10
3x+2y+z=7

Answer 1

Respuesta:

$(x.y.z) = ( \frac{97}{32} . - \frac{19}{32} . - \frac{29}{32} )$

Explicación paso a paso:

$x - 3y + 2z = 3 \\ 2x + y - 5z = 10 \\ r = 7y - 9z = 4 \\ \\ x - 3y + 2z = 3 \\ 3x + 2y + z = 7 \\ r = 11y - 5z = - 2 \\ \\ 7y - 9z = 4 \\ 11y - 5z = - 2 \\ \\ z = ( - \frac{29}{32} ) \\ y = ( - \frac{19}{32} ) \\ \\ x - 3y + 2z = 3 \\ \\x - 3( - \frac{19}{32} ) + 2( - \frac{29}{32} ) = 3 \\ \\ x = \frac{97}{32} \\ \\ x = \frac{97}{32} \\ y = - \frac{19}{32} \\ z = - \frac{29}{32} \\ \\ (x.y.z) = ( \frac{97}{32} . - \frac{19}{32} . - \frac{29}{32} ) \\ \\ \frac{97}{32} - 3( - \frac{19}{32} ) + 2( - \frac{29}{32} ) = 3 \\2( \frac{97}{32}) + ( - \frac{19}{32} ) - 5( - \frac{29}{32} ) =10 \\3( \frac{97}{32} ) + 2 ( - \frac{19}{32} ) + ( - \frac{29}{32} ) =7 \\ \\ 3 = 3 \\ 10 = 10 \\ 7 = 7$

espero que sea lo que estás buscando