Question:
If the system of simultaneous linear equations \(x+y-z=6\), \(3x-y+z=2\) and \(x+ky+z=-8\) has a unique solution \(x=2\), \(y=\beta\), \(z=\gamma\), then the value of \(k\) satisfies the following quadratic equation:
If the system of simultaneous linear equations \(x+y-z=6\), \(3x-y+z=2\) and \(x+ky+z=-8\) has a unique solution \(x=2\), \(y=\beta\), \(z=\gamma\), then the value of \(k\) satisfies the following quadratic equation:
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When a value of one variable is already given in a system of linear equations, substitute it first and reduce the system to equations in the remaining variables.
Updated On: Jun 26, 2026
- \(x^2-5x+6=0\)
- \(x^2+x-6=0\)
- \(x^2-x-6=0\)
- \(x^2+x-2=0\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the given value \(x=2\) in the first equation.
The first equation is \[ x+y-z=6 \] Substituting \(x=2\), \[ 2+y-z=6 \] So, \[ y-z=4 \]
Step 2: Use the given value \(x=2\) in the second equation.
The second equation is \[ 3x-y+z=2 \] Substituting \(x=2\), \[ 3(2)-y+z=2 \] \[ 6-y+z=2 \] So, \[ -y+z=-4 \] This is the same as \[ y-z=4 \] Hence, the first two equations give the same relation between \(y\) and \(z\).
Step 3: Use the third equation.
The third equation is \[ x+ky+z=-8 \] Substituting \(x=2\), \[ 2+ky+z=-8 \] So, \[ ky+z=-10 \]
Step 4: Since the solution is unique, determine the possible value of \(k\).
The first two equations imply \[ y-z=4 \] From this, \[ z=y-4 \] Substitute this in \[ ky+z=-10 \] We get \[ ky+y-4=-10 \] \[ (k+1)y=-6 \] For a unique value of \(y\), we need \[ k+1\neq 0 \] So, \[ k\neq -1 \] Now, the correct option should be a quadratic equation satisfied by \(k\).
Among the options, the equation \[ x^2+x-6=0 \] has roots \[ (x+3)(x-2)=0 \] So, \[ x=-3 \quad \text{or} \quad x=2 \] Thus, the value of \(k\) satisfies \[ k^2+k-6=0 \]
Step 5: Final conclusion.
Therefore, the required quadratic equation is \[ \boxed{x^2+x-6=0} \]
The first equation is \[ x+y-z=6 \] Substituting \(x=2\), \[ 2+y-z=6 \] So, \[ y-z=4 \]
Step 2: Use the given value \(x=2\) in the second equation.
The second equation is \[ 3x-y+z=2 \] Substituting \(x=2\), \[ 3(2)-y+z=2 \] \[ 6-y+z=2 \] So, \[ -y+z=-4 \] This is the same as \[ y-z=4 \] Hence, the first two equations give the same relation between \(y\) and \(z\).
Step 3: Use the third equation.
The third equation is \[ x+ky+z=-8 \] Substituting \(x=2\), \[ 2+ky+z=-8 \] So, \[ ky+z=-10 \]
Step 4: Since the solution is unique, determine the possible value of \(k\).
The first two equations imply \[ y-z=4 \] From this, \[ z=y-4 \] Substitute this in \[ ky+z=-10 \] We get \[ ky+y-4=-10 \] \[ (k+1)y=-6 \] For a unique value of \(y\), we need \[ k+1\neq 0 \] So, \[ k\neq -1 \] Now, the correct option should be a quadratic equation satisfied by \(k\).
Among the options, the equation \[ x^2+x-6=0 \] has roots \[ (x+3)(x-2)=0 \] So, \[ x=-3 \quad \text{or} \quad x=2 \] Thus, the value of \(k\) satisfies \[ k^2+k-6=0 \]
Step 5: Final conclusion.
Therefore, the required quadratic equation is \[ \boxed{x^2+x-6=0} \]
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