Question

For what values of $k$ do all the solutions of the system of equations $x+ky=3,kx+4y=6$ satisfy the conditions $x>1,y>0$?

Answer 1

Given equationns

$x+ky=3\to \left(1\right)$

$kx+4y=6\to \left(2\right)$

Both equations can't be parallel lines

$\therefore \frac{-1}{k}\ne \frac{-k}{4}$

$\therefore k\ne \pm 2$

$k\times \left(1\right)-\left(2\right)\Rightarrow (k^2-4)y=3k-6$

$\therefore y=\frac{3}{k+2}$ $(\because k\ne \pm 2)$

$k\times \left(2\right)-4\times \left(1\right)\Rightarrow (k^2-4)x=6k-12$

$x=\frac{6}{k+2}$

$x>1$ and $y>0$

$\frac{6}{k+2}>1$ and $\frac{3}{k+2}>0$

$k+2<6$ and $k+2>0$

$k<4$ and $k>-2$

$\therefore k\in (-2,4)$