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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 Jacobi 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 Jacobi methodSolution:Total Equations are `2` `2x+5y=21` `x+2y=8` From the above equations `x_(k+1)=1/2(21-5y_(k))` `y_(k+1)=1/2(8-x_(k))` Initial gauss `(x,y) = (0,0)` Solution steps are `1^(st)` Approximation `x_1=1/2[21-5(0)]=1/2[21]=10.5` `y_1=1/2[8-(0)]=1/2[8]=4` `2^(nd)` Approximation `x_2=1/2[21-5(4)]=1/2[1]=0.5` `y_2=1/2[8-(10.5)]=1/2[-2.5]=-1.25` `3^(rd)` Approximation `x_3=1/2[21-5(-1.25)]=1/2[27.25]=13.625` `y_3=1/2[8-(0.5)]=1/2[7.5]=3.75` `4^(th)` Approximation `x_4=1/2[21-5(3.75)]=1/2[2.25]=1.125` `y_4=1/2[8-(13.625)]=1/2[-5.625]=-2.8125` `5^(th)` Approximation `x_5=1/2[21-5(-2.8125)]=1/2[35.0625]=17.5312` `y_5=1/2[8-(1.125)]=1/2[6.875]=3.4375` `6^(th)` Approximation `x_6=1/2[21-5(3.4375)]=1/2[3.8125]=1.9062` `y_6=1/2[8-(17.5312)]=1/2[-9.5312]=-4.7656` `7^(th)` Approximation `x_7=1/2[21-5(-4.7656)]=1/2[44.8281]=22.4141` `y_7=1/2[8-(1.9062)]=1/2[6.0938]=3.0469` Equations are Divergent... Intertions are tabulated as below | Iteration | x | y | | 1 | 10.5 | 4 | | 2 | 0.5 | -1.25 | | 3 | 13.625 | 3.75 | | 4 | 1.125 | -2.8125 | | 5 | 17.5312 | 3.4375 | | 6 | 1.9062 | -4.7656 | | 7 | 22.4141 | 3.0469 |
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