# SOLUTION: 1.evaluate |2x+3|-|x-4|given that x<-2 2.solve the inequality (x^2-1)/(x^2-4)>=0 3.solve the inequality 1/(x-2)-1/(x-1)>=0

Algebra ->  Algebra  -> Absolute-value -> SOLUTION: 1.evaluate |2x+3|-|x-4|given that x<-2 2.solve the inequality (x^2-1)/(x^2-4)>=0 3.solve the inequality 1/(x-2)-1/(x-1)>=0      Log On

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 Question 417567: 1.evaluate |2x+3|-|x-4|given that x<-2 2.solve the inequality (x^2-1)/(x^2-4)>=0 3.solve the inequality 1/(x-2)-1/(x-1)>=0Answer by lwsshak3(6522)   (Show Source): You can put this solution on YOUR website!1.evaluate |2x+3|-|x-4|given that x<-2 2.solve the inequality (x^2-1)/(x^2-4)>=0 3.solve the inequality 1/(x-2)-1/(x-1)>=0 .. 1. |2x+3|-|x-4|=-(2x+3)-(x-4)=-2x-3-x+4=-3x+1 .. 2.(x^2-1)/(x^2-4)>=0 write in factored form, (x+1)(x-1)/(x+2)(x-2)>=0 Critical points: -2, -1, 1, 2 Show critical points on a number line. Using test points,determine whether x-values in the following intervals make the function positive or negative. (-infinity,-2),(-2,-1),(-1,1),(1,2),(2,infinity) Starting from the right side, numbers >2 will make the(2,infinity)interval positive. Moving to the left, signs of the intervals will switch every time we go thru a critical point. (This occurs only because critical points are of multiplicity 1 or odd) Starting from the right,(2,infinity)is positive,(1,2) is negative,,(-1,1)positive,(-2,-1)negative,and (-infinity,-2)positive. Solution:(-infinity,-2] U [-1,1] U [2,infinity). Brackets (]) are used to show that the end points are included in satisfying the function. .. 3.1/(x-2)-1/(x-1)>=0 combine into a single fraction. (x-1)-(x-2)/(x-1)(x-2)>=0 x-1-x+2/(x-1)(x-2)>=0 1/(x-1)(x-2)>=0 critical points are 1 and 2 Put these points on a number line. Following the same procedure as above, Solution: (-infinity,1] U [2, infinity)