course Mth 158
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution:
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
014. `* 14
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Question: * 1.6.12 (was 1.6.6). Explain how you found the real solutions of the equation | 1 - 2 z | + 6 = 9.
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Your solution:
| 1-2 z| + 6 = 9 first add -6 to both sides
1-2 z = 9 – 6
1-2 z = 3 the absolute value of 1-2 z is 3 units either way of 1-2 z so
1-2 z = 3 or 1-2 z = -3
-2z = 3 - 1 -2z = -3 -1
-2z = 2 divide by -2 on both sides -2z = -4
z = -1 z = 2
{-2, 1}
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
{-2/3,2} **
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Self-critique (if necessary):
I don’t understand how the solution set is {-2/3, 2} What did I do wrong?
You were right. That's an error in my editing.
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Self-critique Rating: 3
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Question: * 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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Your solution:
| x^2 +3x - 2 | = 2
x^2 + 3x -2 = 2 or x^2 +3x -2 = -2
x^2 + 3x = 2 + 2 add 2 to both sides x^2 + 3x = -2 + 2
x^2 + 3x = 4 x^2 + 3x = 0
set this side to 0 factor
x^2 + 3x -4 = 0 factor both sides
(x-1)(x+4) = 0 x(x+3) = 0
x = 1 or x = -4 x = 0 or x = -3
{1, -4} {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. **
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Self-critique (if necessary):
I don’t see any discrepancies.
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Self-critique Rating: 3
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Question: * 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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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 add – 4 to all sides
-2-4 < x+4-4 < 2-4
-6 < x < -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
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Self-critique (if necessary):
There are no discrepancies, although 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?
The 2 doesn't get moved.
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 us
x + 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
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Self-critique Rating: 3
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Question: * 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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Your solution:
| -x - 2 | >= 1
-1 >= -x -2 >= 1
2-1 >= -x >= 1+2
1 >= -x >= 3 or
-1 <= x <= -3
{ -infinity, -3} {-1, infinity}
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}. **
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Self-critique (if necessary):
I left off that the first set was united with the second set.
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Self-critique Rating: 3
&#This looks very good. Let me know if you have any questions. &#