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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.)
\n" ); document.write( "Solution 1:
\n" ); document.write( "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\":
\n" ); document.write( "
\n" ); document.write( "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:
\n" ); document.write( "(x+1)/(2x-3) < 2
\n" ); document.write( "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:
\n" ); document.write( "2x-3 > 0 and x+1 < 2(2x-3) or 2x-3 < 0 and x+1 < 2(2x-3)
\n" ); document.write( "(Note: Keep in mind that this is what we get from just the left side of our main (\"colored\") inequality!)
\n" ); document.write( "Now we solve all four of these:
\n" ); document.write( "2x > 3 and x+1 < 4x-6 or 2x < 3 and x+1 < 4x-6
\n" ); document.write( "x > 3/2 and 1 < 3x-6 or x < 3/2 and 1 < 3x-6
\n" ); document.write( "x > 3/2 and 7 < 3x or x < 3/2 and 7 < 3x
\n" ); document.write( "x > 3/2 and 7/3 < x or x < 3/2 and 7/3 < x
\n" ); document.write( "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.
\n" ); document.write( "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
\n" ); document.write( "x > 7/3 or x < 3/2.
\n" ); document.write( "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
\n" ); document.write( "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.
\n" ); document.write( "(x+1)/(2x-3) > -2
\n" ); document.write( "2x-3 > 0 and x+1 > -2(2x-3) or 2x-3 < 0 and x+1 < -2(2x-3)
\n" ); document.write( "2x > 3 and x+1 > -4x+6 or 2x < 3 and x+1 < -4x+6
\n" ); document.write( "x > 3/2 and 5x+1 > 6 or x < 3/2 and 5x+1 < 6
\n" ); document.write( "x > 3/2 and 5x > 5 or x < 3/2 and 5x < 5
\n" ); document.write( "x > 3/2 and x > 1 or x < 3/2 and x < 1
\n" ); document.write( "The two on the left reduce to x > 3/2 and the two on the right reduce to x < 1 leaving:
\n" ); document.write( "x > 3/2 or x < 1
\n" ); document.write( "Now our full solution is:
\n" ); document.write( "x > 7/3 or x < 3/2 and x > 3/2 or x < 1
\n" ); document.write( "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:
\n" ); document.write( "x < 1 or x > 7/3.
\n" ); document.write( "Solution 2:
\n" ); document.write( "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:
\n" ); document.write( "|(x+1)/(2x-3)| < 2
\n" ); document.write( "|(x+1)|/|(2x-3)| < 2
\n" ); document.write( "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:
\n" ); document.write( "|(x+1)| < 2|(2x-3)|
\n" ); document.write( "Since 2 = |2| we can rewrite this as
\n" ); document.write( "|(x+1)| < |2||(2x-3)|
\n" ); document.write( "Now we can use another property of absolute values: |a||b|=|ab| to \"merge\" the absolute values on the right:
\n" ); document.write( "|(x+1)| < |2*(2x-3)|
\n" ); document.write( "|(x+1)| < |(4x-6)|
\n" ); document.write( "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:
\n" ); document.write( "(x+1) < |(4x-6)| and (x+1) > -|(4x-6)|
\n" ); document.write( "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:
\n" ); document.write( "(x+1) < |(4x-6)| and -1(x+1) < |(4x-6)|
\n" ); document.write( "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:
\n" ); document.write( "(x+1) < (4x-6) or -(x+1) > (4x-6) and -1(x+1) < (4x-6) or -1*-1(x+1) > (4x-6)
\n" ); document.write( "Solving all these we get:
\n" ); document.write( "1 < 3x-6 or -x-1 > 4x-6 and -x-1 < 4x-6 or x+1 > 4x-6
\n" ); document.write( "7 < 3x or -1 > 5x-6 and -1 < 5x-6 or 1 > 3x-6
\n" ); document.write( "7/3 < x or 5 > 5x and 5 < 5x or 7 > 3x
\n" ); document.write( "7/3 < x or 1 > x and 1 < x or 7/3 > x
\n" ); document.write( "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:
\n" ); document.write( "x < 1 or x > 7/3
\n" ); document.write( "which happens to be the same solution we got using solution #1!
\n" ); document.write( "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. \n" ); document.write( "