How do you solve the system #-4x-5y-z=18#, #-2x-5y-2z=12#, and #-2x+5y+2z=4#?

Question

5602 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-18T22:00:18

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

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

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

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

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

#"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#