From the give equations
Let
#5x+3y-2z=-12" "#first equation
#x+7y-9z=7" "#second equation
#x-y+z=-4" "#third equation
Also the equations are equivalent to the following; after multiplying the terms of second and third equations by 5
#5x+3y-2z=-12" "#first equation
#5x+35y-45z=35" "#second equation
#5x-5y+5z=-20" "#third equation
Now we eliminate x using the first and second equations by subtraction
#5x+3y-2z=-12" "#first equation
#underline(5x+35y-45z=35" ")#second equation
#0-32y+43z=47#
#32y-43z=47" "#fourth equation
Now we eliminate x using the first and third equations by subtraction
#5x+3y-2z=-12" "#first equation
#underline(5x-5y+5z=-20" ")#third equation
#0+8y-7z=8#
#8y-7z=8" "#fifth equation
Multiply the fifth equation by 4 then subtract the fourth equation
#32y-28z=32" "#fifth equation
#underline(32y-43z=47" ")#fourth equation
#0+15z=-15#
#15z=-15#
#color(red)(z=-1)#
Use the #8y-7z=8" "#fifth equation and #z=-1# to solve for #y#
#8y-7z=8" "#fifth equation
#8y-7(-1)=8" "#fifth equation
#8y+7=8#
#8y=8-7#
#8y=1#
#y=1/8#
#color(red)(y=1/8)#
Use the third equation #x-y+z=-4" "# and #color(red)(z=-1)# and #color(red)(y=1/8)# to solve for #x#
#x-y+z=-4" "#third equation
#x-(1/8)+(-1)=-4" "#third equation
#x=1/8+1-4#
#x=(1-24)/8#
#color(red)(x=-23/8)#
God bless...I hope the explanation is useful.
