********************************************* Question: * 1.6.12 (was 1.6.6). Explain how you found the real solutions of the equation | 1 - 2 z | + 6 = 9. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: | 1 - 2 z | + 6 = 9 1 - 2 z = 3 -2z = 2 z = -1 or 1 - 2 z = -3 -2z = -4 z = 2 {-1, 2} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with | 1-2z| +6 = 9 we add -6 to both sides to get | 1 - 2z| = 3. We then use the fact that | a | = b means that a = b or a = -b: 1-2z=3 or 1-2z= -3 Solving both of these equations: -2z = 2 or -2z = -4 we get z= -1 or z = 2 We express our solution set as {-1, 2} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 1.6.30 (was 1.6.24). Explain how you found the real solutions of the equation | x^2 +3x - 2 | = 2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: | x^2 +3x - 2 | = 2 x^2 +3x - 2 = 2 x^2 + 3x - 4 = 0 (x + 4)(x - 1) = 0 {-4 , 1} or | x^2 +3x - 2 | = -2 x^2 +3x - 2 = -2 x^2 + 3x = 0 x(x + 3) = 0 {0 , -3} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 1.6.40 \ 36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: | x + 4 | + 3 < 5 x + 4 < 2 -2 < x + 4 < 2 -6 < x < -2 (-6 , -2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2 STUDENT QUESTION I was hoping to see more in the given solution as to why we move 2 to the left of the inequality. I think there is a formula for that, but I don’t remember what it is. Could you explain why we move the 2? INSTRUCTOR RESPONSE The 2 doesn't get moved. To understand what's going on: Think about the inequality | A | < = 4. This is clearly true if A = 4, 3, 2, 1 or 0. It's also clearly true if A = -1, -2, -3 or -4. It's not true if A = -5 or -6 or -7, etc.. So | A | < = 4 means the same thing as-4 <= A <= 4. More generally | A | < B says the same thing as - B < A < B. In your solution you said that | x + 4 | + 3 < 5 add -3 to both sides give usx + 4 < 2 This isn't so. The | | signs don't go away when you add -3 to both sides. You get | x + 4 | < 2, which means the same thing as-2 < x + 4 < 2 because of the rule we just say, that | A | < B means -B < A < B. Correcting your solution: | x + 4 | + 3 < 5 add -3 to both sides| x + 4 | < 2 add -2 to the left of the inequality -2 < x + 4 < 2 apply the rule for | A | < B with A = x + 4 and B = 2-2-4 < x+4-4 < 2-4 simplify to get-6 < x < -2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 1.6.52 \ 48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: | -x - 2 | >= 1 -x - 2 >= 1 -x >= 3 x <= -3 or | -x - 2 | <= -1 -x - 2 <= -1 -x <= 1 x >= -1 (-∞ , -3] U [-1 , ∞) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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}. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Is the interval notation in my solution correct?