$3x\:+\:10\:+\:2x\:=\:50$
$\left(3x\right)\:\left(7y\right)$
$\frac{2^3}{2^0}$
$x^2+10x-144$
$\frac{dy}{dx}=\frac{x^2-1}{xy}$
$\frac{dx}{dt}=-2$
$\left(x^2+1\right)y'+\left(4xy\right)=x$
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