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Solve the following system of linear equations in three variables 3x – 2y + z =2 ,2x + 3y – z = 5 ,x + y + z = 6.

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1 Answer

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3x - 2y + z = 2 .....(1)

2x + 3y - z = 5.....(2)

x + y + z = 6 .....(3)

Substituting eqn(6) in eqn(4), we get,

5(1)+y=7[y=2]...........(7)

Substituting eqn(6) and eqn (7) in eqn(3), we get,

1+2+z=6[z=3]...........(8)

Ans:

[x = 1]
[y = 2]
[z = 3]

Verification:

Substituting the values of x,y and z in eqn(2), we get,

LHS = 2(1)+3(2)−3

= 2+6−3=5

= RHS

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