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Question Number 120953 by bemath last updated on 04/Nov/20

$$\mathrm{solve}\mathrm{in}\mathrm{x}\in \mathbb{R}$$$$\mid 3\mathrm{x}-4\mid =\mathrm{x}-5$$

Answered by TANMAY PANACEA last updated on 04/Nov/20

$$when(3x-4)>0\to x>\frac{4}{3}$$$$(3x-4)=x-5$$$$2x=-1$$$$x=\frac{-1}{5}$$$$\mathit{b}\mathit{u}\mathit{t}\mathit{x}=\frac{-1}{5}\mathit{d}\mathit{o}\mathit{e}\mathit{s}\mathit{n}\mathit{o}\mathit{t}\mathit{f}\mathit{o}\mathit{l}\mathit{l}\mathit{o}\mathit{w}\mathit{c}\mathit{o}\mathit{n}\mathit{d}\mathit{i}\mathit{t}\mathit{i}\mathit{o}\mathit{n}$$$$\mathit{x}>\frac{4}{3}\mathit{s}\mathit{o}\mathit{x}=\frac{-1}{5}cannotbeasolution$$$$★now$$$$when(3x-4)<0$$$$x<\frac{4}{3}$$$$-(3x-4)=x-5$$$$-3x-x=-9$$$$4x=9$$$$x=\frac{9}{4}$$$$\mathit{b}\mathit{u}\mathit{t}\mathit{x}=\frac{9}{4}\mathit{d}\mathit{o}\mathit{e}\mathit{s}\mathit{n}\mathit{o}\mathit{t}\mathit{f}\mathit{o}\mathit{l}\mathit{l}\mathit{o}\mathit{w}\mathit{c}\mathit{o}\mathit{n}\mathit{d}\mathit{i}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{x}<\frac{4}{3}$$$$\mathit{S}onosolution$$$$$$$$$$

Answered by ebi last updated on 04/Nov/20

$$\mid 3x-4\mid =x-5$$$$$$$$CaseA$$$$3x-4⩾0$$$$3x-4=x-5$$$$2x=-1\to x=-0.5$$$$\mid 3(-0.5)-4\mid \stackrel{?}{=}(-0.5)-5$$$$\mid -5.5\mid \stackrel{?}{=}-5.5$$$$5.5\ne -5.5$$$$\therefore x=-0.5isnotthesolution.$$$$$$$$CaseB$$$$3x-4<0$$$$3x-4=-(x-5)$$$$4x=9\to x=2.25$$$$\mid 3(2.25)-4\mid \stackrel{?}{=}2.25-5$$$$\mid -2.75\mid \stackrel{?}{=}-2.75$$$$2.75\ne -2.75$$$$\therefore x=2.25isnotthesolution.$$$$nosolution$$

Commented by MJS_new last updated on 04/Nov/20

$$\mid 3\times 2.25-4\mid \ne 2.25-5$$$$2.75\ne -2.75$$

Commented by ebi last updated on 04/Nov/20

$$Yes.Thankyousirfornoticethemistake.$$$$$$

Answered by liberty last updated on 04/Nov/20

$$\mid 3\mathrm{x}-4\mid =\mathrm{x}-5,\mathrm{since}\mid 3\mathrm{x}-4\mid ⩾0\mathrm{for}\mathrm{all}\mathrm{x}ϵ\mathbb{R}$$$$\mathrm{then}\mathrm{it}\mathrm{follow}\mathrm{that}\mathrm{x}⩾5$$$$\mathrm{consider}\mid 3\mathrm{x}-4\mid =\mathrm{x}-5\to \{\begin{array}{l}3\mathrm{x}-4=\mathrm{x}-5\\ \mathrm{or}\\ 3\mathrm{x}-4=5-\mathrm{x}\end{array}$$$$\{\begin{array}{l}2\mathrm{x}=-1\to \mathrm{x}=-\frac{1}{2}\left(\mathrm{rejected}\right)\\ 4\mathrm{x}=9\to \mathrm{x}=\frac{9}{4}\left(\mathrm{rejected}\right)\end{array}$$$$\mathrm{thus}\mathrm{the}\mathrm{equality}\mathrm{has}\mathrm{no}\mathrm{solution}$$$$\mathrm{or}\mathrm{x}=\varnothing .$$

Answered by mathmax by abdo last updated on 04/Nov/20

$$\mid 3\mathrm{x}-4\mid =\mathrm{x}-5\Rightarrow \mid 3\mathrm{x}-4\mid -\mathrm{x}+5=0\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=0$$$$\mathrm{x}-\mathrm{\infty }\frac{4}{3}+\mathrm{\infty }$$$$\mid 3\mathrm{x}-4\mid -3\mathrm{x}+403\mathrm{x}-4$$$$\mathrm{f}\left(\mathrm{x}\right)-4\mathrm{x}+92\mathrm{x}+1$$$$3\mathrm{x}-4-\mathrm{x}+5=2\mathrm{x}+1\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}-4\mathrm{x}+9\mathrm{if}\mathrm{x}⩽\frac{4}{3}\\ 2\mathrm{x}+1\mathrm{if}\mathrm{x}⩾\frac{4}{3}\end{array}$$$$\mathrm{if}\mathrm{x}⩽\frac{4}{3}\mathrm{f}\left(\mathrm{x}\right)=0\Rightarrow -4\mathrm{x}+9=0\Rightarrow \mathrm{x}=\frac{9}{4}\mathrm{but}\frac{9}{4}>\frac{4}{3}\mathrm{its}\mathrm{not}\mathrm{solution}$$$$\mathrm{if}\mathrm{x}⩾\frac{4}{3}\mathrm{f}\left(\mathrm{x}\right)=0\Rightarrow 2\mathrm{x}+1=0\Rightarrow \mathrm{x}=-\frac{1}{2}<\frac{4}{3}\to \mathrm{not}\mathrm{solution}$$$$\mathrm{so}\mathrm{no}\mathrm{solution}\mathrm{for}\mathrm{this}\mathrm{equation}$$

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