course Mth 158
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RESPONSE --> | 1 - 2 z | + 6 = 9 | 1 - 2 z | = 3 1 - 2 z = 3 -2z = 2 z = -1 1 - 2 z = -3 -2z = -4 z = 2 {-1,2}
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18:56:24 ** 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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RESPONSE --> ok
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18:58:49 **** 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 --> | x^2 +3x - 2 | = 2 x^2 +3x - 2 = 2 x^2 +3x - 4 = 0 (x + 4) (x - 1) = 0 x = -4 or x = 1 x^2 + 3x - 2 = -2 x^2 + 3x= 0 x(x + 3) = 0 x = 0 or x = -3 {-4, 1, 0, -3}
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18:58:58 ** 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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RESPONSE --> ok
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19:00:39 **** query 1.6.36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5.
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RESPONSE --> | x + 4 | + 3 < 5 -2 < x + 4 < 2 -6 < x < -2 (-6, -2)
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19:00:46 STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2
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RESPONSE --> ok
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19:04:34 **** query 1.6.48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1.
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RESPONSE --> | -x - 2 | >= 1 -x - 2 <= -1 -x <= 1 x >= -1 [ -1, infinity ) -x - 2 >= 1 -x >= 3 x <= -3 (- infinity, -3]
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19:04:44 **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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RESPONSE --> ok
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