Question

If the system of equations $x+y=1,(c+2)x+(c+4)y=6,(c+2{)}^{2}x+(c+4{)}^{2}y=36$
are consistent, then the value of $c$ can be

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B), (D)

$4$

For system of equations to be consistent:
$\Delta =0$
$\Rightarrow \Delta =\left[\begin{array}{ccc}1& 1& -1\\ (c+2)& (c+4)& -6\\ (c+2{)}^2& (c+4{)}^2& -36\end{array}\right]=0$

By $C_1\to C_1-C_2$ and $C_2\to C_2+C_3$,

$\Rightarrow \left[\begin{array}{ccc}0& 0& -1\\ (-2)& (c-2)& -6\\ -4(c+3)& (c-2)(c+10)& -36\end{array}\right]=0$

Taking $(-2)$ common from $C_1,(c-2)$ common from $C_2$ and $(-1)$ common from $C_3,$ we get
$\Rightarrow (-2)\cdot (c-2)\cdot (-1)\left[\begin{array}{ccc}0& 0& 1\\ 1& 1& 6\\ 2(c+3)& (c+10)& 36\end{array}\right]=0$
$\therefore c=2,4$