course MTH 158 When writing the solution set for an inequality that has an either or solution is it acceptable to write {x | x < 3, x > 10} or do you need to include or {x | x < 3 OR x > 10}?
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15:02:30 **** query 1.6.12 (was 1.6.6). Explain how you found the real solutions of the equation | 1 - 2 z | + 6 = 0.
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RESPONSE --> 12. |1 - 2z| + 6 = 9 |1 - 2z| = 9 - 6 |1 - 2z| = 3 1 - 2z = 3 1 - 2z = -3 -2z = 3 - 1 -2z = -3 - 1 -2z = 2 -2z = -4 z = 2/-2 z = -4/-2 z = -1 z = 2 The solution set to this equation is {-1, 2}.
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15:04:02 11-08-2008 15:04:02 ** Starting with | 1-2z| +6 = 9 we add -6 to both sides to get | 1 - 2z| = 3. We then use the fact that | a | = b means that a = b or a = -b: 1-2z=3 or 1-2z= -3 Solving both of these equations: -2z = 2 or -2z = -4 we get z= -1 or z = 2 We express our solution set as {-2/3,2} **
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NOTES -------> I follow your solution all the way up to the solution set where I think you may have mistyped. : )
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15:06:32 **** query 1.6.30 (was 1.6.24). Explain how you found the real solutions of the equation | x^2 +3x - 2 | = 2
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RESPONSE --> 30. |x^2 + 3x - 2| = 2 x^2 + 3x - 2 = 2 x^2 + 3x - 2 = -2 x^2 + 3x - 2 - 2 = 0 x^2 + 3x - 2 + 2 = 0 x^2 + 3x - 4 = 0 x^2 + 3x = 0 (x - 1)(x + 4) = 0 x(x + 3) = 0 x - 1 = 0 x + 4 = 0 x = 0 x + 3 = 0 x = 1 x = -4 x = 0 x = -3 The solution set to this equation is {-4, -3, 0, 1}.
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15:07:34 11-08-2008 15:07:34 ** My note here might be incorrect. If the equation is | x^2 +3x -2 | = 2 then we have x^2 + 3x - 2 = 2 or x^2 + 3x - 2 = -2. In the first case we get x^2 + 3x - 4 = 0, which factors into (x-1)(x+4) = 0 with solutions x = 1 and x = -4. In the second case we have x^2 + 3x = 0, which factors into x(x+3) = 0, with solutions x = 0 and x = -3. **
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NOTES -------> I hope your solution is correct, it's the same one I got.
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15:09:21 **** query 1.6.40 \ 36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5.
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RESPONSE --> 40. |x + 4| + 3 < 5 |x + 4| < 5 - 3 |x + 4| < 2 x + 4 < 2 x + 4 > -2 x < 2 - 4 x > -2 - 4 x < -2 x > -6 The solution set to this inequality is {x | -6 < x < -2}. Using interval notation (-6, -2).
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15:09:40 11-08-2008 15:09:40 STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2
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NOTES -------> ok
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15:13:27 **** query 1.6.52 \ 48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1.
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RESPONSE --> 52. |x + 2| >= 1 |x + 2| <= -1 x + 2 <= -1 OR x + 2 >= 1 x <= -3 x >= -1 The solution set for this inequality is {x|x <= -3 or x <= -1}. Using interval notation (-infinity, -3/2) U (3/2, infinity).
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15:18:58 11-08-2008 15:18:58 **Correct solution: | -x -2 | >= 1 Since | a | > b means a > b or a < -b (note the word 'or') we have -x-2 >= 1 or -x -2 <= -1. These inequalities are easily solved to get -x >= 3 or -x <= 1 or x <= -3 or x >= -1. So our solution is {-infinity, -3} U {-1, infinity}. **
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NOTES -------> I checked the solution to problem 51 in the solution manual and they divided both absolute values by -1 (at least that's how it appeared to me, it isn't clearly stated) so that's how I worked this problem. However, in their solution they did not change the sign even though they had divided by a negative. Their solution looks like this: |-2x| > |-3| |2x| > 3 2x < -3 or 2x > 3 x < -3/2 or x > 3/2 {x | x < -3/2 or x > 3/2} OR {-infinity, -3/2}U{3/2, infinity) Can you offer any insight as to why the inequality didn't change and if it was supposed to change is this the right solution? I changed the inequality in the assigned problem (52) and achieved the same result as you.
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