How do you solve the following linear system: #3x + 4y - z =1, 3x - y - 4z = 0, x + 3y - 3z = 9#?

2 Answers

Answer:

#x=-138/55#
#y=86/55#
#z=-25/11#

Explanation:

From the given equations
#3x+4y-z=1#first equation
#3x-y-4z=0#second equation
#x+3y-3z=9#third equation

Let us eliminate x first using first and second equations by subtraction

#3x+4y-z=1#first equation
#3x-y-4z=0#second equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#0*x+5y+3z=1#
#5y+3z=1# fourth equation

Let us eliminate x first using first and third equations by subtraction

#3x+4y-z=1#first equation
#x+3y-3z=9#third equation is also

#3x+9y-9z=27#third equation, after multiplying each term by 3

perform subtraction using the new third equation and the first equation

#3x+4y-z=1#first equation
#3x+9y-9z=27#third equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#0*x-5y+8z=-26#
#-5y+8z=-26# fifth equation

Solve for y and z simultaneously using fourth and fifth equations using addition

#5y+3z=1# fourth equation
#-5y+8z=-26# fifth equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#0*y+11z=-25#
#11z=-25#
#z=-25/11#

Solve y using #5y+3z=1# fourth equation and #z=-25/11#

#5y+3z=1# fourth equation
#5y+3(-25/11)=1# fourth equation
#5y-75/11=1#
#5y=75/11+1#
#5y=86/11#
#y=86/55#

Solve for x using #3x+4y-z=1#first equation and #y=86/55# and #z=-25/11#

#3x+4y-z=1#first equation
#3x+4(86/55)-(-25/11)=1#first equation
#3x+344/55+25/11=1#
#3x=1-344/55-25/11#
#3x=(55-344-125)/55#
#3x=-414/55#
#x=-138/55#

Check using the original equations

#3x+4y-z=1#first equation
#3(-138/55)+4(86/55)-(-25/11)=1#first equation

#-414/55+344/55+25/11=1#
#(-414+344+125)/55=1#
#55/55=1#
#1=1#

#3x-y-4z=0#second equation
#3(-138/55)-(86/55)-4(-25/11)=0#second equation
#-414/55-86/55+100/11=0#
#-500/55+100/11=0#
#0=0#

#x+3y-3z=9#third equation
#(-138/55)+3(86/55)-3(-25/11)=9#third equation
#-138/55+258/55+75/11=9#
#120/55+75/11=9#
#24/11+75/11=9#
#99/11=9#
#9=9#

The solution set is
#x=-138/55#
#y=86/55#
#z=-25/11#

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

Mar 18, 2016

Answer:

#(x,y,z)=(-138/55,86/55,-25/11)#

Explanation:

We have the three equations:

#{(3x+4y-z=1" "" "" "" "" ""eq. 1"),(3x-y-4z=0" "" "" "" "" ""eq. 2"),(x+3y-3z=9" "" "" "" "" ""eq. 3"):}#

Multiply #"eq. 3"# by #-3# and add it to #"eq. 1"#:

#{:(3x+4y-z=1),(ul(-3x-9y+9z=-27" "+)),(-5y+8z=-26" "" "" "" "" ""eq. 4"):}#

Multiply #"eq. 1"# by #-3# and add it to #"eq. 3"#:

#{:(-9x-12y+3z=-3),(ul(x+3y-3z=9" "" "" "+)),(-8x-9y=6" "" "" "" "" "" "" ""eq. 5"):}#

Multiply #"eq. 2"# by #3# and add it to #"eq. 3"#:

#{:(9x-3y-12z=0),(ul(x+3y-3z=9" "" "+)),(10x-15z=9" "" "" "" "" "" "" ""eq. 6"):}#

Multiply #"eq. 5"# by #5# and #"eq. 6"# by #4# and add the two:

#{:(-40x-45y=30),(ul(40x-60z=36" "" "+)),(-45y-60z=66" "" "" "" "" "" ""eq. 7"):}#

Multiply #"eq. 4"# by #-9# and add it to #"eq. 7"#:

#{:(45y-72z=234),(ul(-45y-60z=66" "" "+)),(-132z=300" "" "" "" "" "" "" "" ""eq. 8"):}#

From #"eq. 8"#, we can deduce that

#color(red)z=300/(-132)=-150/66=-50/22color(red)(=-25/11#

Use this value of #z# in #"eq. 6"# to solve for #x#:

#10x-15z=9" "=>" "10x-15(-25/11)=9#

#color(white)(sl)=>" "10x+375/11=9#

#color(white)(sl)=>" "10x+375/11=99/11#

#color(white)(sl)=>" "10x=-276/11#

#color(white)(sl)=>" "color(red)x=-276/110color(red)(=-138/55#

We can also use the value of #z# we found to solve for #y# by plugging into #"eq. 4":#

#-5y+8z=-26" "=>" "-5y+8(-25/11)=-26#

#color(white)(sl)=>" "-5y-200/11=-26#

#color(white)(sl)=>" "-5y-200/11=-26#

#color(white)(sl)=>" "-5y-200/11=-286/11#

#color(white)(sl)=>" "-5y=-86/11#

#color(white)(sl)=>" "color(red)(y=86/55#

This gives us the solution set of #(x,y,z)# as:

#color(blue)((-138/55,86/55,-25/11)#