Question

How do you solve the system `x-4y-z=-3`, `x+2y+z=5`, and `3x-7y-2z=-6`?

Answer 1

`x=0.5; y=-0.5; z=5.5`

Explanation:

Add first two expressions to eliminate
the z
`x-4y-z+x+2y+z=-3+5`
`2x-2y=2`

Multiply the first expression by -2, then add to the third expression to eliminate the z again
`-2x+8y+2z=6`
`-2x+8y+2z+3x-7y-2z=6-6`
`x+y=0`

Multiply this answer by 2, then add to the first result to eliminate the y.
`2x+2y=0`
`2x+2y+2x-2y=0+2`
`4x=2`
`x=0.5`

Substitute the x-value into one of the result formulae and solve for y
`0.5+y=0`
`y=-0.5`

Substitute the x-value and the y-value into one of the original formulae and solve for z
`0.5-1+z=5`
`z=5.5`

Substitute your answers into BOTH of the other original formulae to check your results
`0.5+2-5.5=-3. True`
`1.5+3.5-11=-6. True`

Answer 2

`x=1/2`, `y=-1/2` and `z=11/2`

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((1,-4,-1,|,-3),(1,2,1,|,5),(3,-7,-2,|,-6))`

I have written the equations not in the sequence as in the question in order to get `1` as pivot.

Perform the folowing operations on the rows of the matrix

`R2larrR1-R2`; `R3larrR3-3R1`

`A=((1,-4,-1,|,-3),(0,6,2,|,8),(0,5,1,|,3))`

`R2larr(R2)/2`

`A=((1,-4,-1,|,-3),(0,3,1,|,4),(0,5,1,|,3))`

`R2larrR2-R3`

`A=((1,-4,-1,|,-3),(0,-2,0,|,1),(0,5,1,|,3))`

`R2larr(R2)/(-2)`

`A=((1,-4,-1,|,-3),(0,1,0,|,-1/2),(0,5,1,|,3))`

`R1larrR1+4R2; R3-5R2`

`A=((1,0,-1,|,-5),(0,1,0,|,-1/2),(0,0,1,|,11/2))`

`R1larrR1+R3`

`A=((1,0,0,|,1/2),(0,1,0,|,-1/2),(0,0,1,|,11/2))`

Thus, `x=1/2`, `y=-1/2` and `z=11/2`