$-3x\le-2x+8$

$y'\:=\:-y\:-\:2x,\:y\:\left(0\right)\:=\:1$

$\int tan^5\left(\frac{x}{4}\right)\cdot sec^2\left(\frac{x}{4}\right)$

$\frac{\cos\left(x\right)+\cos\left(y\right)}{\sin\left(x+y\right).\sin\left(x-y\right)}$

$\int\frac{lny}{y^4}dy$

$2x+9>3x$

$\frac{4}{x}-3\le\frac{2}{x}-7$