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Click here to Find the value of h,k for which the system of equations has a Unique or Infinite or no solution calculator
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Solution
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Solution provided by AtoZmath.com
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Solving systems of linear equations using Gauss Elimination Back Substitution method calculator
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 1. `2x+y+z=5,3x+5y+2z=15,2x+y+4z=8`
2. `2x+5y=16,3x+y=11`
3. `2x+5y=21,x+2y=8`
4. `2x+y=8,x+2y=1`
5. `2x+3y-z=5,3x+2y+z=10,x-5y+3z=0`
6. `x+y+z=3,2x-y-z=3,x-y+z=9`
7. `x+y+z=7,x+2y+2z=13,x+3y+z=13`
8. `2x-y+3z=1,-3x+4y-5z=0,x+3y-6z=0`
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Example1. Solve Equations 2x+5y=21,x+2y=8 using Gauss Elimination Back Substitution methodSolution:Total Equations are `2` `2x+5y=21 -> (1)` `x+2y=8 -> (2)` Converting given equations into matrix form `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` | |
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`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`
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