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Question Number 101363 by bemath last updated on 02/Jul/20

$$\mathrm{The}\mathrm{solution}\mathrm{set}\mathrm{of}\mathrm{inequality}$$$$\frac{\sqrt{{(3\mathrm{x}-7)}^2}-2}{\mathrm{x}-3}⩽\frac{3-\sqrt{{\mathrm{x}}^2}}{\mathrm{x}-3}\mathrm{is}\mathrm{\_}\mathrm{\_}$$$$\left(\mathrm{A}\right)(-\mathrm{\infty },\frac{1}{2}]\left(\mathrm{D}\right)[\frac{1}{2},\mathrm{\infty })$$$$\left(\mathrm{B}\right)[\frac{1}{2},1]\left(\mathrm{E}\right)(-\mathrm{\infty },\frac{2}{3}]$$$$\left(\mathrm{C}\right)(-\mathrm{\infty },1]$$

Answered by 1549442205 last updated on 02/Jul/20

$$\iff \frac{\mid 3\mathrm{x}-7\mid -2}{\mathrm{x}-3}-\frac{3-\mid \mathrm{x}\mid }{\mathrm{x}-3}⩽0\left(1\right)$$$$\mathrm{i})\mathrm{if}\mathrm{x}⩾\frac{7}{3}\mathrm{then}\left(1\right)\iff \frac{3\mathrm{x}-7-2-(3-\mathrm{x})}{\mathrm{x}-3}⩽0$$$$\iff \frac{4\mathrm{x}-12}{\mathrm{x}-3}⩽0\iff \frac{4(\mathrm{x}-3)}{\mathrm{x}-3}⩽0\iff 4⩽0$$$$\Rightarrow \left(1\right)\mathrm{has}\mathrm{no}\mathrm{solutions}$$$$\mathrm{ii})\mathrm{if}0⩽\mathrm{x}<\frac{7}{3}\mathrm{then}\left(1\right)\iff \frac{7-3\mathrm{x}-2-(3-\mathrm{x})}{\mathrm{x}-3}$$$$\iff \frac{2-2\mathrm{x}}{\mathrm{x}-3}⩽0\iff \frac{1-\mathrm{x}}{\mathrm{x}-3}⩽0\iff \mathrm{x}\in \{(-\mathrm{\infty };1]\cup (3;+\mathrm{\infty })\}\cap [0;\frac{7}{3})$$$$\iff \mathrm{x}\in [0;1]$$$$\mathrm{iii})\mathrm{if}\mathrm{x}<0\mathrm{then}\left(1\right)\iff \frac{7-3\mathrm{x}-2-(3+\mathrm{x})}{\mathrm{x}-3}⩽0$$$$\iff \frac{2-4\mathrm{x}}{\mathrm{x}-3}⩽0\iff \frac{1-2\mathrm{x}}{\mathrm{x}-3}⩽0\iff (-\mathrm{\infty };\frac{1}{2}]\cup (3;+\mathrm{\infty })\cap (-\mathrm{\infty };0)$$$$\iff (-\mathrm{\infty };0)$$$$\mathrm{Combinating}\mathrm{three}\mathrm{above}\mathrm{cases}\mathrm{we}\mathrm{get}$$$$\mathbf{The}\mathbf{solutions}\mathbf{of}\mathbf{given}\mathbf{inequality}\mathbf{is}$$$$\mathbf{x}\in (-\mathrm{\infty };1],\mathbf{so}\mathbf{choose}\mathbf{answer}\mathbf{C}$$

Commented by 1549442205 last updated on 02/Jul/20

$$\mathrm{Thank}\mathrm{you}.\mathrm{You}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

Commented by bemath last updated on 02/Jul/20

$$\mathrm{thank}\mathrm{you}.\mathrm{agree}$$

Commented by Rasheed.Sindhi last updated on 02/Jul/20

$$\iff \frac{\mid 3\mathrm{x}-7\mid -2}{\mathrm{x}-3}-\frac{3-\mid \mathrm{x}\mid }{\mathrm{x}-3}⩽0\left(1\right)$$$$\mathrm{i})\mathrm{if}\mathrm{x}⩾\frac{7}{3}\mathrm{then}\left(1\right)\iff \frac{3\mathrm{x}-7-2-(3-\mathrm{x})}{\mathrm{x}-3}⩽0$$$$\iff \frac{4\mathrm{x}-12}{\mathrm{x}-3}⩽0\iff \frac{4(\mathrm{x}-3)}{\mathrm{x}-3}⩽0\iff 4⩽0$$$$Atx=\frac{7}{3},\frac{4(\mathrm{x}-3)}{\mathrm{x}-3}⩽0\nRightarrow 4⩽0$$$$Butx=\frac{7}{3},\frac{4(\mathrm{x}-3)}{\mathrm{x}-3}⩽0\nRightarrow 4⩾0$$$$\because \mathrm{x}-3<0$$$$\Rightarrow \left(1\right)\mathrm{has}\mathrm{no}\mathrm{solutions}$$$$\mathrm{ii})\mathrm{if}0⩽\mathrm{x}<\frac{7}{3}\mathrm{then}\left(1\right)\iff \frac{7-3\mathrm{x}-2-(3-\mathrm{x})}{\mathrm{x}-3}$$$$.....................$$$$..........$$$$.....$$

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