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A solution set of the equations x+2y+z=1...

A solution set of the equations `x+2y+z=1`, `x+3y+4z=k`, `x+5y+10z=k^(2)` is

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The correct Answer is:
K=1,2
k=1,x=5t+1,y=-3t,z=t `t in R`
k=2,x=5t-1,y=1-3t,z=t
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Knowledge Check

  • The number of distinct real values of K such that the system of equations x+2y+z=1 , x+3y+4z=K, x+5y+10z = K^2 has infinitely many solutions is :

    A
    0
    B
    4
    C
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    D
    3

  • The equations x + 2y + 3z = 1, x − y + 4z = 0 and 2x + y + 7z = 1has

    A
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    B
    only two solutions
    C
    no solution
    D
    infinitely many solutions

  • For the equations x+2y+3z=1,2x+y+3z=2 and 5x+5y+9z=4 ,

    A
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    B
    there exists infinitely many solutions
    C
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    D
    none of the above

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    A solution set of the equations x-3y-8z+10=0,3x+y-4z=0, 2x+5y+6z=13 is (A)(2K+1,3-2K,K) (B) (2K-1,3+2K,K) (C) (1-2K,3-2K,K) (D) (2K-1,3-2K,K)

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