Question 212810
APPLY BASIC DEFINITION OF ABSOLUTE VALUES TO GET THE POSSIBLE SOLUTION SETS
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by the basic definition of absolute values, |(x+1)/(2x-3)| < 2 becomes:
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{{{(x+1)/(2x-3) < 2}}} when the expression within the absolute value signs is positive.
and:
{{{- ((x+1)/(2x-3)) < 2}}} when the expression within the absolute value signs is negative.
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When you multiply {{{-((x+1)/(2x-3)) < 2}}} by -1 on both sides of the equation, it becomes:
{{{((x+1)/(2x-3)) > -2)}}} when the expression within the absolute value signs is negative.
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POSSIBLE SOLUTION SETS
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You have 2 possible solution sets.
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set 1 is when the expression within the absolute value signs is positive and is:
{{{(x+1)/(2x-3) < 2}}}
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set 2 is when the expression within the absolute value signs is negative and is:
{{{(x+1)/(2x-3) > -2}}}
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DETERMINE WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGN IS POSITIVE AND WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGN IS NEGATIVE.
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The numerator is positive when {{{x + 1 >= 0}}} which becomes when {{{x >= -1}}}
The denominator is positive when {{{2x - 3 >= 0}}} which becomes when {{{x >= 3/2}}}.
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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.
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Numerator is >= 0 when x >= -1
Denominator is >= 0 when x > 3/2 (= 3/2 not allowed).
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The expression is positive when both the numerator and the denominator are both positive.  This happens when {{{x > 3/2}}}.
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The expression is also positive when both the numerator and the denominator are both negative.
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this happens when {{{x < -1}}}.
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The expression is positive when:
{{{x > 3/2}}} or {{{x < -1}}}
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This means the expression is negative when:
{{{x < 3/2}}} and {{{x >= -1}}}
This can be written as:
{{{-1 <= x < 3/2}}}
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remember that x cannot be equal to 3/2.
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SOLVE FOR WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGNS IS POSITIVE.
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When the expression within the absolute sign is positive, the possible solution set is:
{{{(x+1)/(2x-3) < 2}}}
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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.
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If the denominator is positive, then you would solve this is as follows:
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Equation to solve is:
{{{(x+1)/(2x-3) < 2}}}
Multiply both sides of this equation by (2x-3) to get:
{{{x+1 < 2 * (2x-3)}}}
Remove parentheses to get:
{{{x+1 < 4x - 6}}}
subtract 1 from both sides and subtract 3x from both sides to get:
{{{-3x < -7}}}
divide both sides by 3 and multiply both sides by -1 to get:
{{{x > 7/3}}}
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If the denominator is negative, then you would solve this as a follows:
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Equation to solve is:
{{{(x+1)/(2x-3) < 2}}}
Multiply both sides of this equation by (2x-3) to get:
{{{x+1 > 2 * (2x-3)}}}
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Note that the inequality reversed because we were multiplying by a negative number which is represented by the expression (2x-3) in the denominator.
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Remove parentheses to get:
{{{x + 1 > 4x - 6}}}
subtract 4x from both sides of the equation and subtract 1 from both sides of the equation to get:
{{{-3x > -7}}}
multiply both sides of this equation by -1 and divide both sides of this equation by 3 to get:
{{{x < 7/3}}}
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You have 2 possible solutions when the expression within the absolute value sign is positive.
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Those 2 possible solutions are:
{{{x > 7/3}}} when the denominator is positive
{{{x < 7/3}}} when the denominator is negative
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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.
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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
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Your possible solutions for when the absolute value expression is positive are:
{{{x > 7/3}}} or {{{x < -1}}}
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SOLVE FOR WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGNS IS NEGATIVE
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When the expression within the absolute value signs is negative, the possible solution set is:
{{{(x+1)/(2x-3) > -2}}}
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The boundaries for this occur when:
{{{x < 3/2}}} and {{{x >= -1}}}
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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.
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When the denominator is negative you would solve as follows:
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Equation to solve is:
{{{(x+1)/(2x-3) > -2}}}
Multiply both sides of this equation by (2x-3) to get:
{{{x+1 < -2 * (2x-3)}}}
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Note that (2x-3) is a negative number so multiplying both sides of this equation by a negative number reverses the inequality.
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Remove parentheses to get:
{{{x + 1 < -4x + 6}}}
Subtract 1 from both sides of this equation and add 4x to both sides of this equation to get:
{{{5x < 5}}}
divide both sides of this equation by 5 to get:
{{{x < 1}}} when the denominator is negative.
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Since the solution set is negative when {{{x >= -1}}} and {{{x < 3/2}}}, then the possible solution when the expression within the absolute value sign is negative is:
{{{x >= -1}}} and {{{x < 1}}}
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this can be written as:
{{{-1 <= x < 1}}}
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PUT ALL THE INFORMATION TOGETHER SO YOU CAN ANALYZE IT TO CONFIRM THAT THE ANSWER IS GOOD.
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When the expression within the absolute value signs is positive, the possible solutions are:
{{{x > 7/3}}} or {{{x < -1}}}
When the expression within the absolute value signs is negative, the possible solutions are:
{{{-1 <= x < 1}}}
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These 2 solutions can be combined to show as:
{{{x < 1}}} or {{{x > 7/3}}}
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.
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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)
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When x = -2:
{{{abs((x+1)/(2x-3))}}} becomes .14 < 2 so this is good as it should be.
When x = -1:
{{{abs((x+1)/(2x-3))}}} becomes 0 < 2 so this is good as it should be.
When x = 0:
{{{abs((x+1)/(2x-3))}}} becomes .33 so this is good as it should be.
When x = 1:
{{{abs((x+1)/(2x-3))}}} becomes 2 NOT < 2 so this is NOT good as it should be.
When x = 7/3:
{{{abs((x+1)/(2x-3))}}} becomes 2 so this is NOT good as it should be.
When x = 8/3:
{{{abs((x+1)/(2x-3))}}} becomes 1.57 so this is good as it should be.
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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.
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http://theo.x10hosting.com/
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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.
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