Question

If the system of equations $x+y-3=0,(1+K)x+(2+K)y-8=0$ & $x-(1+K)y+(2+K)=0$ is consistent then the value of K may be

• A. 1
• B. $\frac{3}{5}$
• C. $-\frac{5}{3}$
• D. 2

Answer 1

Answer: (A), (C)

1
$-\frac{5}{3}$

Given that, system of equations $x+y-3=0,(1+k)x+(2+k)y-8=0$ & $x-(1+k)y+(2+k)=0$ is consistent.
treating the given system as, $3$ equations in $3$ variables.
Augumented matrix is $\left[\begin{array}{cccc}1& 1& 0& 3\\ 1+k& 2+k& 0& 8\\ 1& -1-k& 0& -2-k\end{array}\right]$
above system is consistent if $\left|\begin{array}{ccc}1& 1& 3\\ 1+k& 2+k& 8\\ 1& -1-k& -2-k\end{array}\right|=0$
$\Rightarrow (-k^2+4k+4)+(k62+3k+10)-3(k^2+3k+3)=0$
$\Rightarrow 3k^2+2k-5=(k-1)(3k+5)=0$
$\therefore k=1$ or $k=\frac{-5}{3}$
Hence, options A and C.