Question

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

Answer 1

`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.

Answer 2

`(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)`