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Home > Matrix & Vector calculators > Solving systems of linear equations using Gauss Elimination Back Substitution method example
4. Gauss Elimination Back Substitution method example ( Enter your problem )
( Enter your problem )
  1. Example `2x+5y=21,x+2y=8`
  2. Example `2x+5y=16,3x+y=11`
  3. Example `2x+3y-z=5,3x+2y+z=10,x-5y+3z=0`
  4. Example `x+y+z=3,2x-y-z=3,x-y+z=9`
 		Other related methods
  1. Inverse Matrix method
  2. Cramer's Rule method
  3. Gauss-Jordan Elimination method
  4. Gauss Elimination Back Substitution method
  5. Gauss Seidel method
  6. Gauss Jacobi method
  7. Elimination method
  8. LU decomposition using Gauss Elimination method
  9. LU decomposition using Doolittle's method
  10. LU decomposition using Crout's method
  11. Cholesky decomposition method
  12. SOR (Successive over-relaxation) method
  13. Relaxation method
3. Gauss-Jordan Elimination method
(Previous method)		2. Example `2x+5y=16,3x+y=11`
(Next example)

1. Example `2x+5y=21,x+2y=8`


1. Solve Equations 2x+5y=21,x+2y=8 using Gauss Elimination Back Substitution method

Solution:
Total Equations are `2`

`2x+5y=21 -> (1)`

`x+2y=8 -> (2)`

Converting given equations into matrix form
`2``5``21`
`1``2``8`


`R_2 larr R_2-1/2xx R_1`

 = 
`2``5``21`
 `0` `0=1-1/2xx2`
`R_2 larr R_2-1/2xx R_1`		 `-1/2` `-1/2=2-1/2xx5`
`R_2 larr R_2-1/2xx R_1`		 `-5/2` `-5/2=8-1/2xx21`
`R_2 larr R_2-1/2xx R_1`


`i.e.`

`2x+5y=21 ->(1)`

`-1/2y=-5/2 ->(2)`

Now use back substitution method
From (2)
`-1/2y=-5/2`

`=>y=-5/2xx-2=5`

From (1)
`2x+5y=21`

`=>2x+5(5)=21`

`=>2x+25=21`

`=>2x=21-25=-4`

`=>x=(-4)/(2)=-2`

Solution using back substitution method.
`x = -2`

`y = 5`

This material is intended as a summary. Use your textbook for detail explanation.
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3. Gauss-Jordan Elimination method
(Previous method)		2. Example `2x+5y=16,3x+y=11`
(Next example)


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