(D) x = 3 .10 Solution of | x + 3| > 2x – 1 is 2.16 (C) x = 3 (D) x 2 2.17

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( |x+3|>|2 x-1| ) ( begin{aligned}|x+3| &left{begin{array}{rr}x+3 & x geq-3 -x-3 & x leq 3end{array}right.|2 x-1| &=left{begin{array}{cl}2 x-1 & x geq 1 / 2 -2 x+1 & x leq 1 / 2end{array}right.end{aligned} ) (D) ( 16 x in(-infty,-3] ) ( |x+3|=-x-3 quad|2 x-1|=-2 x+1 ) ( -x-3>-2 x+1 ) ( x-4>0 ) ( x>4 ) Novalued (2) ( 1 /-x in[-3,1 / 2] ) ( |x+3|=x+3 quad|2 x-| mid=-2 x+1 ) ( x+3>-2 x+1 ) ( 3 x->-2 quad x>-2 / 3 ) ( Rightarrow x in(-2 / 31 / 2] ) (5)( left(f-x in[1 / 2, infty)^{3}right. ) ( |x+3|=x+3 quad|2 x-1|=2 x-1 ) ( x+3>2 x-1 ) ( Rightarrow x<4 ) ( Rightarrow x in[1 / 2,4) ) ( Rightarrow x in(-2 / 3,4) )

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