How do you solve the system -4x-5y-z=184x5yz=18, -2x-5y-2z=122x5y2z=12, and -2x+5y+2z=42x+5y+2z=4?

1 Answer
Mar 18, 2017

{x=-4,y=0,z=-2}{x=4,y=0,z=2}

Explanation:

-4x-5y-z=18" , "(1)4x5yz=18 , (1)

-2x-5y-2z=12" , "(2)2x5y2z=12 , (2)

-2x+5y+2z=4" , "(3)2x+5y+2z=4 , (3)

"let us sum the equations numbered as (2) and (3) so as to eliminate"let us sum the equations numbered as (2) and (3) so as to eliminate

"the terms named 'y' and 'z'".the terms named 'y' and 'z'.

-2xcancel(-5y)cancel(-2z)-2xcancel(+5y)cancel(+2z)=12+4

"rearrange the equation above."

-2x-2x=12+4

-4x=16" , "x=-4

"let us write as x=4 in the equation numbered as (1)"

-4(-4)-5y-z=18

16-5y-z=18

-5y-z=18-16

-5y-z=2" , "(4)

"let us write as x=4 in the equation numbered as (2)"

-2(-4)-5y-2z=12

8-5y-2z=12

-5y-2z=12-8

-5y-2z=4" , "(5)

"now , subtract (5) from (4) "

-5y-z-(-5y-2z)=2-4

cancel(-5y)-z cancel(+5y)+2z=-2

-z+2z=-2

z=-2"

"now use (4)"

-5y-(-2)=2

-5y+2=2

-5y=0

y=0