NTSE Stage 2 SAT 2016 Question Paper

$\mathrm{x}-\mathrm{y}=2$ ...(1)
$\mathrm{kx}+\mathrm{y}=3$ ...(2)
by adding (1) and (2)
$\mathrm{kx}+\mathrm{x}=5$
$\mathrm{x}\left(\mathrm{k}+1\right)=5$
$\mathrm{x}=\frac{5}{\mathrm{k}+1}$
putting value of x in equation (1), we get
$\frac{5}{\mathrm{k}+1}-\mathrm{y}=2$
$\frac{5}{\mathrm{k}+1}-2=\mathrm{y}$
$\frac{5-2\mathrm{k}-2}{\mathrm{k}+1}=\mathrm{y}$
$\mathrm{y}=\frac{3-2\mathrm{k}}{\mathrm{k}+1}$
y should be positive as they intersect in 1st quadrant therefore
$\mathrm{y}>0$

 +    -    +
∴ k should lie between – 1 and $\frac{3}{2}$