How do you solve #-3x-y=0, y-4z=15, x-2y+z=-10#?

1 Answer
Jan 1, 2017

Answer:

#x=-1"; "y=3"; "z=-3#

Explanation:

# ((x,y,z,|,-),(-3,-1,0,|,0),(0,1,-4,|,15),(1,-2,1,|,-10)) #
#color(brown)(" "R_3+2R_2)#
#color(brown)(" "darr)#

# ((-3,-1,0,|,0),(0,1,-4,|,15),(1,0,-7,|,+20)) #
#color(brown)(" "R_1+3R_3)#
#color(brown)(" "darr)#

# ((0,-1,-21,|,60),(0,1,-4,|,15),(1,0,-7,|,+20)) #
#color(brown)(" "R_1+R_2)#
#color(brown)(" "darr)#

# ((0,0,-25,|,75),(0,1,-4,|,15),(1,0,-7,|,+20)) #
#color(brown)(" "R_1-:(-25))#
#color(brown)(" "darr)#

# ((0,0,1,|,-3),(0,1,-4,|,15),(1,0,-7,|,+20)) #
#color(brown)(" "R_2+4R_1#
#color(brown)(" "darr)#

# ((0,0,1,|,-3),(0,1,0,|,3),(1,0,-7,|,+20)) #
#color(brown)(" "R_3+7R_1#
#color(brown)(" "darr)#

# ((0,0,1,|,-3),(0,1,0,|,3),(1,0,0,|,-1)) #
#color(brown)(" "R_3+7R_1#

#x=-1"; "y=3"; "z=-3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("check")#

#-3x-y=0" " ->" " -3(-1)-3= 0 " "color(red)(larr"True")#

#y-4z=15" " ->" "3-4(-3)=15" "color(red)(larr"True")#

#x-2y+z=-10" "->""-1-2(3)-3= -10" "color(red)(larr"True")#