Linear InequalitiesDate: 11/26/97 at 08:06:54 From: Maura Subject: Linear Inequalities How do I solve the problem (x-2)/(x+1)<(x-4)/(x-1) ? Date: 11/26/97 at 14:04:30 From: Doctor Rob Subject: Re: Linear Inequalities Ahh! Good question! First I would bring both terms over to one side of the equation, and combine them: (x-2)/(x+1) - (x-4)/(x-1) < 0, [(x-2)*(x-1)-(x+1)*(x-4)]/[(x+1)*(x-1)] < 0, [x^2-3*x+2-x^2+3*x+4]/[(x+1)*(x-1)] < 0, 6/[(x+1)*(x-1)] < 0 Now I would divide both sides by 6 to simplify, and take the reciprocal of both sides (since 1/a < 0 if and only if a < 0): (x+1)*(x-1) < 0 The roots of the polynomial on the left are -1 and 1, which divide the real number line into three regions: 1. x < -1 2. -1 < x < 1 3. 1 < x The sign of the polynomial will be the same throughout each region, but it may differ from region to region. To figure out the sign in each region, it suffices to pick a representative point in each, such as -1000, 0, and +1000, and figure out the sign at those points. The union of those regions where the sign is negative forms your solution set. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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