Solve the following inequations: $3 \mathrm{x}-36>0$
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Answer
$ 3 x-36>0 $ Adding 36 both sides, we get $ \begin{aligned} & 3 x-36+36>0+36 \\ & \therefore 3 x>36 \end{aligned} $ Dividing both sides by 3 , we get $ \begin{aligned} & \frac{3 x}{3}>\frac{36}{3} \\ & \therefore x>12 \\ & \therefore x \text { takes all real values more than } 12 \\ & \therefore \text { Solution set }=(12, \infty) \end{aligned} $ $\therefore x$ takes all real values more than 12 $\therefore$ Solution set $=(12, \infty)$
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