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

##### 2 Answers

#### Answer:

#### Explanation:

From the given equations

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

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

perform subtraction using the new third equation and the first equation

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

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Solve y using

Solve for x using

Check using the original equations

The solution set is

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

#### Answer:

#### 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

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

Multiply

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

Multiply

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

Multiply

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

Multiply

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

From

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

Use this value of

#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

#-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

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