What is the solution of the following system?:#5 x + 3y + -2z = -12, x + 7y - 9z = 7, x + -y + z = -4#

1 Answer

Answer:

#x=-23/8# and #y=1/8# and #z=-1#

Explanation:

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.