Algebra

$x+y>5$ and $x-y>3$ .

Multiplying both the equations, we get:

$\implies\space x^2-y^2=5\times3=15$ . Now, we know that $y^2$ must be at least $1$ . To find minimum value of $x$ , let us take $y^2=1$ . So, we have:

$\therefore\space x^2=15+y^2=15+1=16\space \implies\space x>4$ .

x+y>5 x-y>3 then adding, 2x>8 therefore, x>4