find-the-range-or-ranges-of-value-x-can-take-for-x-6-gt-2x-3-

Question Number 10777 by j.masanja06@gmail.com last updated on 24/Feb/17

$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{or}\:\left(\mathrm{ranges}\right)\:\mathrm{of}\:\mathrm{value}\:\mathrm{x} \\ $$$$\mathrm{can}\:\mathrm{take}\:\mathrm{for}\:\:\mathrm{x}+\mathrm{6}>\left[\mathrm{2x}+\mathrm{3}\right] \\ $$

Commented by mrW1 last updated on 24/Feb/17

$${do}\:{you}\:{mean}\:{x}+\mathrm{6}>\mid\mathrm{2}{x}+\mathrm{3}\mid\:? \\ $$

Commented by j.masanja06@gmail.com last updated on 25/Feb/17

$$\mathrm{yes}\:! \\ $$

Answered by mrW1 last updated on 24/Feb/17

$${if}\:\mathrm{2}{x}+\mathrm{3}\geqslant\mathrm{0},\:{i}.{e}.\:{x}\geqslant−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}+\mathrm{6}>\mathrm{2}{x}+\mathrm{3} \\ $$$${x}<\mathrm{3} \\ $$$$\Rightarrow−\frac{\mathrm{3}}{\mathrm{2}}\leqslant{x}<\mathrm{3} \\ $$$$ \\ $$$${if}\:\mathrm{2}{x}+\mathrm{3}<\mathrm{0},\:{i}.{e}.\:{x}<−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}+\mathrm{6}>−\mathrm{2}{x}−\mathrm{3} \\ $$$$\mathrm{3}{x}>−\mathrm{9} \\ $$$${x}>−\mathrm{3} \\ $$$$\Rightarrow−\mathrm{3}<{x}<−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{range}\:{of}\:{x}: \\ $$$$−\mathrm{3}<{x}<\mathrm{3} \\ $$

Answered by lee last updated on 25/Feb/17

$$\left.{sol}\:\mathrm{1}\right)\left({x}+\mathrm{6}\right)^{\mathrm{2}} >\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{27}<\mathrm{0},\:−\mathrm{3}<{x}<\mathrm{3} \\ $$$$\left.{sol}\:\mathrm{2}\right) \\ $$$$\left.{i}\right){x}\geqq−\frac{\mathrm{3}}{\mathrm{2}},\:{x}+\mathrm{6}>\mathrm{2}{x}+\mathrm{3} \\ $$$$−\frac{\mathrm{3}}{\mathrm{2}}\leqq{x}<\mathrm{3} \\ $$$$\left.{ii}\right){x}<−\frac{\mathrm{3}}{\mathrm{2}},\:{x}+\mathrm{6}>−\mathrm{2}{x}−\mathrm{3} \\ $$$$−\mathrm{3}<{x}<−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left.\therefore\left.\:\:{i}\right)\cup{ii}\right)\rightarrow−\mathrm{3}<{x}<\mathrm{3} \\ $$

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