Question

The number of values of $k$ for which the system of equations $x+y=2$, $kx+y=4$, $x+ky=5$ has only one solution is _________.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

$x+y=\mathrm{2....}\left(1\right)$

$kx+y=\mathrm{4....}\left(2\right)$

$x+ky=\mathrm{5....}\left(3\right)$

Adding $2$ and $3$, we get

$(k+1)(x+y)=9$

But we know that $x+y=2$

$\therefore k+1$ must be equal to $\frac{9}{2}$

$\therefore$ For the system of equation to have atleast one solution, $k=3.5$

Hence, only $1$ value.