Question

Solve the following inequality:
$\sqrt{3x-10}>\sqrt{6-x}$

Answer 1

$\sqrt{3x-10}>\sqrt{6-x}$

$3x-10>6-x3x+x>16$

$4x>16$

$x>4$

$x\epsilon (4,\infty )-\left(i\right)$

$but(6-x)\&(3x-10)cannotbenegative$

$6-x>0$

$x<6x\epsilon (-\infty ,6)-\left(ii\right)$

$3x-10>0$

$x>\frac{10}{3}x\epsilon (\frac{10}{3},\infty )-\left(iii\right)$

$x\epsilon (4,6)from\left(i\right),\left(ii\right)\&\left(iii\right)$