$10x+4x+4+8x-2$
$5\left(x-2\right)+\left(-3x\right)-4$
$-3x-5=-20$
$2x-8<4x$
$-6-5x-6x=16$
$\frac{x^2}{4}+2\frac{x^2}{6}+\frac{x^2}{9}$
$\frac{h}{6}-1=3$
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