 # Help on a few compound inequality problems?(how to do it) Anonymous
timer Asked: Jul 14th, 2014
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### Question Description

1) -3<2x-1<7

2) -2x+7>3 or 3x-4≥5

3) -1<1/2x<1

4) -4d>8 and 2d>-6

OSTutors
School: New York University   1) Add +1 on both the sides, (-3+1)<2x-1+1<(7+1) => -2<2x<8 ; Divide by 2 on both sides, (-2/2)<(2x/2)<(8/2) => -1<x<4

2)   -2x+7>3 => -2x> 3-7 (subtracting 7 on both sides) => -2x>-4 => -x>-2 (dividing by 2 on both sides) , so x<2 (applying - sides on both sides, so becomes less than) .

3x-4>=5      =>        3x-4+4>=5+4      =>    3x>=9    =>   x>=3 (dividing by 3 on both sides) ;

Hence from 1 and 2 equations  x<2 or x>=3 .  (*****this is or so 'x' can't lie in between them).

3) -1<1/2x<1 =>   -2<(1/)x<2 (multiplying by 2 on both sides)  =>  -(1/2) > x > (1/2)  (less than becomes greater than due to reciprocation).  Hence    (1/2)<x<(-1/2)

4)  -4d>8   => -d>4 (dividing by 4 on both sides) => d<-4 (greater than becomes less than due to change of sign, with multiplication of - on both sides).  AND   similarly 2d>-6  => d>-3  (divide both sides by 2)

Hence   -3<d<-4 (since AND, d lies in between them).

THANK YOU,

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