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Solve `|3x-2|le(1)/(2),x inR.` |
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Answer» We have, `|x|lea iff-alexlea`. `therefore|3x-2|le(1)/(2)iff-(1)/(2)le3x-2le(1)/(2)` `iff(-1)/(2)le3x-2and 3x-2le(1)/(2)` `iff(-1)/(2)+2le3xand3xle(1)/(2)+2` `iff(3)/(2)le3x and 3xle(5)/(2)` `iff(1)/(2)lex and xle(5)/(6)` `(1)/(2)lexle(5)/(6)`. `therefore` solution set `={x inR:(1)/(2)lexle(5)/(6)}=[(1)/(2),(5)/(6)]`. |
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