Question

A value of $c$ for which the system of equations

$x+y=1$

$(c+2)x+(c+4)y=6$

$(c+2{)}^{2}x+(c+4{)}^{2}y=36$ is solvable (consistent) is

• A. $1$
• B. $2$
• C. $4$
• D. none of these

Answer 1

Answer: (B), (C)

The correct options are
B $2$
C $4$

Let $x+y=1$

$(c+2)x+(c+4)y=6$

$(c+2{)}^{2}x+(c+4{)}^{2}y=36$

Then

$\left(\begin{array}{ccc}1& 1& -1\\ (c+2)& (c+4)& -6\\ (c+2{)}^2& (c+4{)}^2& -36\end{array}\right)$

By $C_1=C_1-C_2$ and $C_2=C_2-C_3$,

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

$=-1(-2(c-2\left)\right(c+10)+4(c+3\left)\right(c-2\left)\right)$

$=-1(c-2)(-2(c+10)+4(c+3\left)\right)$

$=-1(c-2)(-2c-20+4c+12)$

$=-2(c-2)(c-4)$

So, $c$ is $2$ or $4$