How do you solve #-2x - 4y + 2z = -8#, #-x - 5y +12z = 5#, #3x + 5y - z = 10#?

1 Answer

Answer:

#x=-9# and #y=8# and #z=3#

Explanation:

From the given equations
#-2x-4y+2z=-8" " "#first equation
#-x-5y+12z=5" " "#second equation
#3x+5y-z=10" " "#third equation

We eliminate the variable x first

multiply the terms of the second equation by 2 then subtract it from the first equation

#-2x-4y+2z=-8" " "#first equation
#-2x-10y+24z=10" " "#second equation
after subtraction
#6y-22z=-18# fourth equation

multiply the original second equation by 3 then add to the third equation

#-3x-15y+36z=15" " "#second equation
#3x+5y-z=10" " "#third equation

after addition
#-10y+35z=25" " "# fifth equation reducible to
#-2y+7z=5larr#reduced fifth equation
Multiply this fifth equation by 3 then add to the fourth equation

#-6y+21z=15" " "#fifth equation
#6y-22z=-18# fourth equation

the result is
#-z=-3#
and
#z=3#

Using the reduced fifth equation #-2y+7z=5# and #z=3# solve for y

#-2y+7z=5#
#-2y+7(3)=5#
#-2y+21=5#
#-2y=-16#
#y=8#

Using the original second equation #-x-5y+12z=5" "#and #y=8# and #z=3# solve for #x#

#-x-5y+12z=5" "#
#-x-5(8)+12(3)=5" "#
#-x-40+36=5#
#-x=9#
#x=-9#

the solution is #x=-9# and #y=8# and #z=3#

God bless...I hope the explanation is useful..