$-x+y=6$
$\left(10x^2\:-x\:+11\right)^2$
$6x+3\le x-9$
$\frac{dy}{dx}=y\cos\left(x\right)+y\sin\left(x\right)$
$\int\left(2x^2+3\right)\left(2x-2\right)dx$
$6\left(4a-3b\right)$
$\frac{5x}{2}^2$
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