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.