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:

• A. \(x^2-5x+6=0\)
• B. \(x^2+x-6=0\)
• C. \(x^2-x-6=0\)
• D. \(x^2+x-2=0\)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Substitute x = 2 into all three equations.
Eq1: \(2+y-z=6\Rightarrow y-z=4\). Eq2: \(6-y+z=2\Rightarrow y-z=4\) (consistent). Eq3: \(2+k y+z=-8\Rightarrow ky+z=-10\).

Step 2: Solve for y and z, then find k.
From \(y-z=4\) and \(ky+z=-10\): adding gives \(y(1+k)=-6\Rightarrow y=\frac{-6}{1+k}\) and \(z=y-4\). For a unique solution the system must be consistent; testing the answer options: if \(k=-3\) or \(k=2\), both satisfy \(k^2+k-6=0\), i.e., \((k+3)(k-2)=0\). So the quadratic is \(x^2+x-6=0\). \[ \boxed{x^2+x-6=0} \]