Solve Equations x+y+z=3,2x-y-z=3,x-y+z=9 using Elimination method
Solution:
Total Equations are `3`
`x+y+z=3 -> (1)`
`2x-y-z=3 -> (2)`
`x-y+z=9 -> (3)`
Select the equations `(1)` and `(2)`, and eliminate the variable `y`.
`x+y+z=3` | ` xx 1->` | | `` | `x` | `+` | `y` | `+` | `z` | `=` | `3` | `` |
| | + | |
`2x-y-z=3` | ` xx 1->` | | `` | `2x` | `-` | `y` | `-` | `z` | `=` | `3` | `` |
| | |
|
| | | `` | `3x` | | | | | `=` | `6` | ` -> (4)` |
Select the equations `(1)` and `(3)`, and eliminate the variable `y`.
`x+y+z=3` | ` xx 1->` | | `` | `x` | `+` | `y` | `+` | `z` | `=` | `3` | `` |
| | + | |
`x-y+z=9` | ` xx 1->` | | `` | `x` | `-` | `y` | `+` | `z` | `=` | `9` | `` |
| | |
|
| | | `` | `2x` | | | `+` | `2z` | `=` | `12` | ` -> (5)` |
Now use back substitution method
From (4)
`3x=6`
`=>x=(6)/(3)=2`
From (5)
`2x+2z=12`
`=>2(2)+2z=12`
`=>2z+4=12`
`=>2z=12-4=8`
`=>z=(8)/(2)=4`
From (1)
`x+y+z=3`
`=>1(2)+1y+1(4)=3`
`=>1y+6=3`
`=>1y=3-6=-3`
Solution using Elimination method.
`x=2,y=-3,z=4`
This material is intended as a summary. Use your textbook for detail explanation.
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