SOLUTION: Please help me solve this equation: (x-1)/(x+2) > 0

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Question 135881: Please help me solve this equation: (x-1)/(x+2) > 0

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
First let's get your terminology correct. %28x-1%29%2F%28x%2B2%29+%3E+0 is NOT an equation, it is an inequality.

You have already completed step 1 for solving rational inequalities, and that is to arrange the inequality so that you have 0 on one side (generally the right) and everything else on the other side.

Step 2: Find the boundary points:

Step 2A: Set the numerator equal to zero and solve: x-1=0 => x=1

Step 2B: Set the denominator equal to zero and solve: x%2B2 => x=-2

Step 3: Determine whether your solution will be inclusive or exclusive of the boundary points.

Step 3A: x=1, since the entire fraction must be larger and not equal to zero, the boundary point x=1 must be excluded from the solution set because if x=1 the numerator is equal to zero so the entire fraction would be zero.

Step 3B: x=-2, since a denominator can never be zero, -2 must be excluded from the solution set.

Step 4: Now you have defined three intervals: (-infinity,-2), (-2,1), and (1,infinity)

Step 5: Select any convenient value from the first interval, -3 will do. Substitute this value into the original inequality:

%28-3-1%29%2F%28-3%2B2%29+=-4%2F-1=4 Since the result is positive, values in this interval are included in the solution set of the inequality. A positive result means the interval is included because the original inequality has the rational expression greater than zero, or positive.

Step 6 and 7: Repeat step 5 for the other two intervals:

Select 0 from the second interval:
%280-1%29%2F%280%2B2%29+=-1%2F2 Since the result is negative, values in this interval are excluded.

Select 2 from the third interval:
%282-1%29%2F%282%2B2%29+=1%2F4 Again, since the result is positive, values in this interval are included in the solution set of the inequality.

Therefore your solution set is all real x such that x%3C2 or x%3E1