Question 217047
Replace a with x and your equations to solve are:
|x-12| < 9 AND |x+2| > 17
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I do this because I'm so used to working with x that I always write x even when I know it's a or b or whatever.
Converting to x eliminates the problems IF I remember to convert back at the end.
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Pure laziness on my part because I already started using x and didn't want to have to go back and convert them to a in all the equations.
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The operating principle is:
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The absolute value of an expression equals the expression if the expression is greater than or equal to 0.
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The absolute value of an expression equals minus the expression if the expression is less than or equal to 0.
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let E be the expression.
Let the equation be |E| > A
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If E >= 0, we remove the absolute value sign and get E > A.
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If E < 0, we remove the absolute value sign and set -E > A.
We multiply both sides of this equation by (-1) to get E < -A.
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We need to find when the expressions within the absolute value signs are positive and when the expressions within the absolute value signs are negative.
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You have a maximum of 4 possible cases.
case 1 is when both expressions are positive within the absolute value signs.
case 2 is when both expressions are negative within the absolute value signs.
case 3 is when the first expression is positive within the absolute value signs, and the second expression is negative within the absolute value signs.
case 4 is when the first expression is negative within the absolute value signs, and the second expression is positive within the absolute value signs.
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We test each expression separately to find out when the expression within the absolute value signs is positive and when the expression within the absolute value signs is negative.
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Take |x-12| < 9
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The expression within the absolute value sign is (x-12).  This is our first expression.
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x-12 >= 0 when x >= 12
x-12 < 0 when x < 12
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The first expression is positive when x >= 12
The first expression is negative when x < 12
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Now take |x+2| > 17
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The expression within the absolute value sign is (x+2). This is our second expression.
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x+2 >= 0 when x >= -2
x+2 < 0 when x < -2
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The second expression is positive when x >= -2.
The second expression is negative when x < -2
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Our 4 cases reduce to 3.
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They are:
Case 1:
Both expressions are negative when x < -2.
Case 2:
Both expressions are positive when x >= 12.
Case 3:
The first expression is negative when x < 12 and the second expression is positive when x >= -2
Case 4:
We do not have a case when first expression is positive and second expression is negative so this one cancels out.
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We look at each of these 3 cases separately.
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case 1 (both expressions are positive within the absolute value signs).
This happens when x >= 12.
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|x-12| < 9 becomes (x-12) < 9 which becomes x < 21.
|x+2| > 17 becoming (x+2) > 17 which becomes x > 15.
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We have 2 possible solutions.
They are:
x < 21 or x > 15.
Since case 1 is when x >= 12, both these solutions are possible.
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We'll move to case number 2 which is when both expressions are negative within the absolute value signs.
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Case 2 (both expressions are negative within the absolute value signs).
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This happens when x < -2
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|x-12| < 9 becomes -(x-12) < 9 becomes (x-12) > -9 when you multiply both sides of this equation by (-1).  This becomes x > 3
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|x+2| > 17 becomes -(x+2) > 17 which becomes (x+2) < -17 when you multiply both sides of this equation by (-1).  This becomes x < -19
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Our two possible solutions are:
x > 3 or x < -19
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Since case 2 is when x < -2, the only possible solution is x < -19.
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We'll move on to case 3 and see what we get.
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Case 3 is: (x-12) is negative and (x+2) is positive.
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This happens when x >= 2 and x < 12
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Since (x-12) is negative within the absolute value signs, we get:
|x-12| < 9 becomes -(x-12) < 9 becomes (x-12) > -9 becomes x > 3.
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Since (x+2) > 17 is positive within the absolute value signs, we get:
|x+2| > 17 becomes (x+2) > 17 becomes x > 15.
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We have two possible solutions of x > 15 and x > 3
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Since Case 2 is when x >= 2 ANDE x < 12, the only possible solution is x > 3 because x > 15 is outside the range of permissible values for Case 2.
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We have 3 possible solutions for this set of equations.
They are:
x > 15 and x < 21
x < -19
x > 3
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We look at each in turn.
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Our equations are:
|x-12| < 9
|x+2| > 17
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When x = -20 < -19, we get:
|x-12| = |-20-12| = |-32| = 32 NOT < 9.
This solution is not good because it does not satisfy one of the equations.
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When x = 4 > 3, we get:
|x-12| = |4-12| = |-8| = 8 < 9 which is OK so we check the second equation.
|x+2| = |4+2| = |6| = 6 NOT > 17.
This solution is not good because it does not satisfy one of the equations.
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When x = 16 > 15 and < 21, we get:
|x-12| = |16-12| = |4| = 4< 9 OK.
|x+2| = |16+2| = |18| = 18 > 17 OK.
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So far, when x > 15 and < 21, we can solve both equations simultaneously.
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We tested the lower limit of 16 for this possible solution set, so we test  the upper limit of these solutions by testing when x = 20.
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When x = 20 > 15 < 21, we get:
|x-12| = |20-12| = 8 < 9 which is good.
|x+2| = |20+2| = 22 > 17 which is also good.
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Within the limits we are ok.
We test outside the limits next to see if the limits were expressed correctly.
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When x = 21, we get:
|x-12| = 9 NOT < 9 which is not good.  The limits on the high end are correct.
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When x = 15, we get:
|x-12| = 3 < 9 which is good.
|x+2| > 17 = 17 NOT > 17 which is not good.  The limits on the low end are correct.
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Before we answer the question, we set x = a to convert back to the original variable name.
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Our answer to the question is:
15 < a < 21
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Those values of a will satisfy both equations of:
|a-12| < 9
and
|a+2| > 17
simultaneously.
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