Question

For how many values of $k$ does the following system of equations have at least one solution?
$x+y=1$; $kx+y=3$; $x+ky=5$

• A. $0$
• B. $1$
• C. $2$
• D. Infinitely many

Answer 1

The correct option is B $1$

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

$kx+y=\mathrm{3....}\left(2\right)$

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

Adding $2$ and $3$, we get

$(k+1)(x+y)=8$

But we know that $x+y=1$

$\therefore k+1$ must be equal to $8$

$\therefore$ For the system of equation to have atleast one solution, $k=7$, hence, only $1$ value.