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

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

Commented by mrW1 last updated on 24/Feb/17

$$doyoumeanx+6>\mid 2x+3\mid ?$$

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

$$\mathrm{yes}!$$

Answered by mrW1 last updated on 24/Feb/17

$$if2x+3⩾0,i.e.x⩾-\frac{3}{2}$$$$x+6>2x+3$$$$x<3$$$$\Rightarrow -\frac{3}{2}⩽x<3$$$$$$$$if2x+3<0,i.e.x<-\frac{3}{2}$$$$x+6>-2x-3$$$$3x>-9$$$$x>-3$$$$\Rightarrow -3<x<-\frac{3}{2}$$$$$$$$\Rightarrow rangeofx:$$$$-3<x<3$$

Answered by lee last updated on 25/Feb/17

$$sol1){(x+6)}^2>{(2x+3)}^2$$$$3x^2-27<0,-3<x<3$$$$sol2)$$$$i)x\geqq -\frac{3}{2},x+6>2x+3$$$$-\frac{3}{2}\leqq x<3$$$$ii)x<-\frac{3}{2},x+6>-2x-3$$$$-3<x<-\frac{3}{2}$$$$\therefore i)\cup ii)\to -3<x<3$$

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