# SOLUTION: What steps would I take to solve the absolute value of |(x+1)/(2x-3)|<2?

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 Question 212810: What steps would I take to solve the absolute value of |(x+1)/(2x-3)|<2? Found 2 solutions by jsmallt9, Theo:Answer by jsmallt9(3757)   (Show Source): You can put this solution on YOUR website! Two solutions are provided. The first solution requires an understanding that when you multiply both sides of an inequality by a negative number, the inequality reverses. The second solution, which follows, requires an understanding of a) properties of absolute values: |a/b| = |a|/|b|; |a||b| = |ab| and b) A good understanding of how to remove the absolute values correctly. Choose whichever makes the most sense. (I hope they both make sense.) Solution 1: It helps to keep in mind what absolute value stands for. The absolute value of some expression represents the distance of the value of the expression from zero on the number line. In your case, |(x+1)/(2x-3)| < 2, says that value of (x+1)/(2x-3) is less than 2 from zero. In other "words": Now we have two inequalities without absolute value. We now solve each one individually. This, as I think you will see, turns out to be the tricky part of this problem. We'll start with the left side: (x+1)/(2x-3) < 2 What we want to do is multiply both sides by 2x-3. This will eliminate the fraction giving a relatively simple inequality to solve. In general, when we multiply both sides of an inequality by a negative we have to reverse the inequality. However, we don't know if 2x-3 will be positive or negative!? And since it makes a difference, we have to account for both possibilities. If 2x-3 is positive, 2x-3 > 0, then we do not reverse the inequality. If 2x-3 is negative, 2x-3 < 0, then we need to reverse the inequality. So the proper way to multiply both sides of our inequality by 2x-3 is: 2x-3 > 0 and x+1 < 2(2x-3) or 2x-3 < 0 and x+1 < 2(2x-3) (Note: Keep in mind that this is what we get from just the left side of our main ("colored") inequality!) Now we solve all four of these: 2x > 3 and x+1 < 4x-6 or 2x < 3 and x+1 < 4x-6 x > 3/2 and 1 < 3x-6 or x < 3/2 and 1 < 3x-6 x > 3/2 and 7 < 3x or x < 3/2 and 7 < 3x x > 3/2 and 7/3 < x or x < 3/2 and 7/3 < x The left two say x > 3/2 and x > 7/3. With some thought we can see that if x > 7/3 then it would have to be greater than 3/2 also. So these two can be reduced to simply x > 7/3. The right two say x < 3/2 and x < 7/3. With similar logic we can determine that if x < 3/2, then it would have to be less than 7/3, too. So these two can be reduced to x < 3/2. Now we have x > 7/3 or x < 3/2. Remember, this is just "half" of our solution. Now we apply the same procedures to the right "half" of our main ("colored") inequality: (x+1)/(2x-3) > -2 To save time and space I will leave out most of the commentary. All the steps are the same and are used for the same reasons as above. (x+1)/(2x-3) > -2 2x-3 > 0 and x+1 > -2(2x-3) or 2x-3 < 0 and x+1 < -2(2x-3) 2x > 3 and x+1 > -4x+6 or 2x < 3 and x+1 < -4x+6 x > 3/2 and 5x+1 > 6 or x < 3/2 and 5x+1 < 6 x > 3/2 and 5x > 5 or x < 3/2 and 5x < 5 x > 3/2 and x > 1 or x < 3/2 and x < 1 The two on the left reduce to x > 3/2 and the two on the right reduce to x < 1 leaving: x > 3/2 or x < 1 Now our full solution is: x > 7/3 or x < 3/2 and x > 3/2 or x < 1 The easiest way to see the solution set is to graph the left pair on one number line and the right pair on another number line and then, since there is an "and" between the halves, see where the two graphs overlap. Unfortunately Algebra.com's software does not graph inequalities so you'll have to do this yourself. The overlapping parts show that the "final" solution is: x < 1 or x > 7/3. Solution 2: A basic property of absolute values is: |a/b| = |a|/|b|. We can use this to split the absolute value of the fraction into a fraction of absolute values: |(x+1)/(2x-3)| < 2 |(x+1)|/|(2x-3)| < 2 Now we can multiply both sides by |(2x-3|. This will eliminate the fraction and, since absolute values are always positive we do not need to be concerned with the possibility of reversing the inequality (like we did in solution 1)! So now we have: |(x+1)| < 2|(2x-3)| Since 2 = |2| we can rewrite this as |(x+1)| < |2||(2x-3)| Now we can use another property of absolute values: |a||b|=|ab| to "merge" the absolute values on the right: |(x+1)| < |2*(2x-3)| |(x+1)| < |(4x-6)| Now we remove the absolute values, one at a time. We'll start on the left. From the perspective of the left side, this is a "less than". In general, "less than" absolute values, |a| < b, can be "removed"/converted to: a < b and a > -b. They are always "and"-type. Using this on our problem we get: (x+1) < |(4x-6)| and (x+1) > -|(4x-6)| Before we remove the second absolute value we need to remove the minus sign in front of the absolute value on the right side. So we multiply both sides by -1, and since we're multiplying be a negative, we need to reverse the inequality: (x+1) < |(4x-6)| and -1(x+1) < |(4x-6)| To remove the absolute value from the right sides we need to look at the inequalities from the perspective of the right side. From the right side these inequalities are both "greater than" inequalities! (It is very helpful to be able to read inequalities from right-to-left!!) "Greater than" absolute value inequalities can be "removed"/converted in the following way: |a| > b becomes a > b or a < -b. "Greater than" absolute values are always "or"-type inequalities. Using this on our inequalities we get: (x+1) < (4x-6) or -(x+1) > (4x-6) and -1(x+1) < (4x-6) or -1*-1(x+1) > (4x-6) Solving all these we get: 1 < 3x-6 or -x-1 > 4x-6 and -x-1 < 4x-6 or x+1 > 4x-6 7 < 3x or -1 > 5x-6 and -1 < 5x-6 or 1 > 3x-6 7/3 < x or 5 > 5x and 5 < 5x or 7 > 3x 7/3 < x or 1 > x and 1 < x or 7/3 > x Again it helps to graph each "half" on a number line and see where the graphs overlap. If we do so we find the overlap ocurrs for: x < 1 or x > 7/3 which happens to be the same solution we got using solution #1! Final note: Notice how, as I solved the simple inequalities I made sure the last coefficient of x was positive. This is a good habit because if we allow the coefficient to become negative then we will have to divide by a negative at some point. And this means we have to remember to reverse the inequality eventually. It's just simpler if we avoid having to remember ans use the "reverse the inequality" rule as much as possible. Answer by Theo(5548)   (Show Source): You can put this solution on YOUR website! APPLY BASIC DEFINITION OF ABSOLUTE VALUES TO GET THE POSSIBLE SOLUTION SETS ----- by the basic definition of absolute values, |(x+1)/(2x-3)| < 2 becomes: ----- when the expression within the absolute value signs is positive. and: when the expression within the absolute value signs is negative. ----- When you multiply by -1 on both sides of the equation, it becomes: when the expression within the absolute value signs is negative. ----- POSSIBLE SOLUTION SETS ----- You have 2 possible solution sets. ----- set 1 is when the expression within the absolute value signs is positive and is: ---- set 2 is when the expression within the absolute value signs is negative and is: ----- DETERMINE WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGN IS POSITIVE AND WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGN IS NEGATIVE. ----- The numerator is positive when which becomes when The denominator is positive when which becomes when . ----- Note that x cannot be equal to 3/2 because then the denominator in the equation is equal to 0 which is not a real number and therefore can't be part of any solution to this problem. ----- Numerator is >= 0 when x >= -1 Denominator is >= 0 when x > 3/2 (= 3/2 not allowed). ----- The expression is positive when both the numerator and the denominator are both positive. This happens when . ----- The expression is also positive when both the numerator and the denominator are both negative. ----- this happens when . ----- The expression is positive when: or ----- This means the expression is negative when: and This can be written as: ----- remember that x cannot be equal to 3/2. ----- SOLVE FOR WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGNS IS POSITIVE. ----- When the expression within the absolute sign is positive, the possible solution set is: ----- You have two possible scenarios within this solution set. The first scenario is when the denominator is positive. The second scenario is when the denominator is negative. ----- If the denominator is positive, then you would solve this is as follows: ----- Equation to solve is: Multiply both sides of this equation by (2x-3) to get: Remove parentheses to get: subtract 1 from both sides and subtract 3x from both sides to get: divide both sides by 3 and multiply both sides by -1 to get: ----- If the denominator is negative, then you would solve this as a follows: ----- Equation to solve is: Multiply both sides of this equation by (2x-3) to get: ----- Note that the inequality reversed because we were multiplying by a negative number which is represented by the expression (2x-3) in the denominator. ----- Remove parentheses to get: subtract 4x from both sides of the equation and subtract 1 from both sides of the equation to get: multiply both sides of this equation by -1 and divide both sides of this equation by 3 to get: ----- You have 2 possible solutions when the expression within the absolute value sign is positive. ----- Those 2 possible solutions are: when the denominator is positive when the denominator is negative ----- When the expression is positive and the denominator is positive, x has to be greater than 3/2, so x > 7/3 does not have to be modified and can stand as is. ----- When the expression is positive and the denominator is negative, x has to be less than -1 so x < 7/3 has to be modified to say x < -1 ----- Your possible solutions for when the absolute value expression is positive are: or ----- SOLVE FOR WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGNS IS NEGATIVE --- When the expression within the absolute value signs is negative, the possible solution set is: ----- The boundaries for this occur when: and ----- The denominator of 2x-3 is negative throughout this boundary so we only have to solve for when the denominator is negative and do not have to solve for when the denominator is positive. ----- When the denominator is negative you would solve as follows: ----- Equation to solve is: Multiply both sides of this equation by (2x-3) to get: ----- Note that (2x-3) is a negative number so multiplying both sides of this equation by a negative number reverses the inequality. ----- Remove parentheses to get: Subtract 1 from both sides of this equation and add 4x to both sides of this equation to get: divide both sides of this equation by 5 to get: when the denominator is negative. ----- Since the solution set is negative when and , then the possible solution when the expression within the absolute value sign is negative is: and ----- this can be written as: ----- PUT ALL THE INFORMATION TOGETHER SO YOU CAN ANALYZE IT TO CONFIRM THAT THE ANSWER IS GOOD. ----- When the expression within the absolute value signs is positive, the possible solutions are: or When the expression within the absolute value signs is negative, the possible solutions are: ----- These 2 solutions can be combined to show as: or We test the intervals to see if they are accurate by taking values within and without those intervals to see if the equation is true or not. ----- Test values for x will be: -2 (should be good) -1 (should be good) 0 (should be good) 1 (should not be good) 7/3 (should not be good) 8/3 (should be good) ----- When x = -2: becomes .14 < 2 so this is good as it should be. When x = -1: becomes 0 < 2 so this is good as it should be. When x = 0: becomes .33 so this is good as it should be. When x = 1: becomes 2 NOT < 2 so this is NOT good as it should be. When x = 7/3: becomes 2 so this is NOT good as it should be. When x = 8/3: becomes 1.57 so this is good as it should be. ----- If you wish to see what the intervals look like, then go to the following website and click on problem number 212810. If it's not there when you look, then wait 30 minutes and try again. It will be there. ----- http://theo.x10hosting.com/ ----- I must thank you for presenting this problem. It expanded my knowledge on how to solve these types of problems. I never tried one before where there was a numerator and denominator in the equation. It was an eye opener to say the least. I wrote a tutorial on how to solve absolute value equations and will definitely place this problem in there as an example of a more complicated type of problem. I'm satisfied that I have the right solution now, but it took a while before I was able to get this problem under control enough to say that. Thanks again. -----