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YOU are better than YOU think. Show
yourself how:
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Read logic
chapters 1 to 5 in online volume Three
Skills for Algebra for greater skills & confidence
in work
and study.
Read notes and textbooks like a lawyer, so that no nuance, no subtlety and no clause escapes your attention. |
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Logic
chapters 1 to 5 re- appear not in sequence, as is or longer,
in Volume 1A, Pattern Based
Reason, Bon Appetite.
Logic
Mastery
Amazing, Amusing, Amorous, Delicious, Delightful, Edifying,
Strengthening Elixir.
It eases work & learning difficulties Makes the hard easier. Opens eyes.
Leads to greater precision.
in reading and
writing
Logic
mastery makes the hard, easier. Logic
mastery leads to better, stronger and richer comprehension. Logic
mastery improves reading and writing. Logic
mastery ease learning difficulties. Logic
mastery gives a headstart. In sum, logic
mastery will develops critical thinking, improve reading and writing,
and give a firmer base for work and studies at many levels. Good luck.
After logic,
(a) continue reading Three
Skills for Algebra, chapters 8 to 14 and do so alongside site area on solving
liinear Equations ; or (b) see this calculus
starter lesson and Volume 3, Why
Slopes & More Math, chapters 2 to 6;
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Caution: Site advice is approximately
correct, for some circumstances, not all. That leaves room for thought |
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What may be learnt and when depends on how skills
and concepts are developed. Making the hard easier and clearer will allow
earlier & richer development of skills and concepts.
Try the Twiddla
Whiteboard. In principle, it allows
to people to draw and chat together online on a copy of this webpage or a clean
sheet. The chat may be via text or audio. Visit www.twiddla.com
to set up whiteboards to work with the webpage of your choice.
For online automated help in senior high school maths & calculus,
visit quickmath.com For Automatic
Calculus and Algebra Help with derivatives, integrals, graphs, linear equations,
matrix algebra, visit calc101.com
With overlap, each site quickmath
& calc101offers a different range of
services, some free, some not, all based on webmathematica. Good luck.
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Absolute Value
The magnitude or absolute value function (a computation rule)
The absolute value or magnitude of a positive number 5 = +5 is the number
itself.
The absolute value or magnitude of zero 0 is zero 0.
The absolute value or magnitude of a negative number _7
is +7. The same result can be obtained by multiplying by -1 or
computing the negative of _ 7 as 7 =
(-1)( _ 7) = - ( _ 7)
The absolute value of a real number gives its distance to 0.
Exercise: Find the distance to zero (absolute value) of the
following numbers: 5.6, 23.85, 3.14, p,
-p, 10, -10, -45.67, 45.67
The computation of the absolute value of a number x can be described in
words, the absolute value of x, that is,
|x| is given by the number x when x > 0 and by
-x = (-1)x when x < 0
or equivalent by the shorthand formula
|
|x| = { |
x when x > 0, and |
| -x iwhen x < 0. |
Some text may write if in place of when.
The first difficulty in dealing with the absolute value function f(x) = |x|
lies in accepting the shorthand formula as the starting point for future
computations and reasoning with absolute value.
Example I: Let us compute | -5|. Now x = -5 < 0. So
according to the formula |x| = -x = -(-5) and the latter gives 5, the same
result as dropping the sign in front of the number).
Example II: Let us compute |+2.5|. Here x = 2.5 >
0. So according to the formula |x| = x = 2.5 and we are done.
Solving Equations with Absolute Value
Example I: Find the solution set S of the
equation |x| =5. Answer: S = {+5, -5} or x =
(+/-) 5
Example II: Find the solution set S of the
equation |Q| =5. Answer: S = {+5, -5} or Q =
(+/-) 5 (the letter used in the equation does not matter)
Example III: Find the solution set S of the
equation |x-3 | = 13.
Answer: Put P = x -3. The equation says |P| = 13 and
hence the P = (+/-)13. So we know that
P = -13 or P = +13.
Since P is shorthand for x - 3, these two case imply respectively that
x -3 = -13 or x - 3 = 13
and hence x = -13 + 3 = -10 or x = 13 + 3 = 16.
So we suspect the solution set S = {-10, 16) or equivalently x = 3 +
(+/-)13. That completes the answer except for a check - checks are always
needed for multi-step answers or processes.
Check: When x = 16, x-3 = 13 and |x-3| = 13
and when x = -10, x-3 = -10 -3 = -13 and |x-3| = 13 again. So our answer
work.
Remark: Our values for x are the only answers as we followed chains
of reason which implied if |x-3| = 13 then x = -10 or x = 16.
Example IV: Find the solution set S of the
equation |x-3 | < 13.
Answer: Put P = x -3. The equation says |P| < 13 and
hence the P belongs to the interval (-13, 13) with end points included
(why?)
Since P is shorthand for x - 3, we require x - 3 belong to the interval -13
to 13 (endpoints excluded). That gives two inequalities to satisfy,
namely
x -3 > -13 and x - 3 < 13
The first requires x > -13+3 = -10 while the second requires x
< 13 + 3 =16.
So we must have -10 < x < 16. So the solution set S =
(-10, 16).
Check: If - 10 < x < 16, we may subtract 3 to find
-13 < x - 3 < 13 and hence |x-3| < 13.
Remark: The endpoints of the solution interval -10 < x < 16
would be included if we were solving |x-3 | < 13 instead of |x-3 |
< 13. Do you see why.
Example V: Find the solution set S of the equation
( | x+3 | - 4 )(x+8) = 0.
Answer: The only way ( | x+3 | - 4 )(x+8) = 0 is if at least
one of the factors is zero. So we must have
( | x+3 | - 4 )= 0 or (x+8) =0
The second factor x+8 = 0 when and only when x = -8. So our first solution
is x = -8.
The first factor is zero when and only when |x+3| = 4 or equivalently x+3 =
(+/-)4 or x = - 3+ (+/-)4. So the first factor is zero when and only
when x = -3+4 = 1 or x = -3 - 4 = -7.
So the first factor is zero for x = 1 or x = - 7 while the second
factor is zero for x = -8.
So in the absence of mistakes in our reasoning, any solution of the ( | x+3
| - 4 )(x+8) = 0 must be an element of the set S = {1, -7, 8}
Now complete the solution by checking that these values work.
Example VI: Find the solution set S of the
equation | x2-17 | = 8.
Solution: x2-17 = (+/-)8. Therefore x2
= (+/-)8. +17 belongs to { 17 - 8, 17+ 8} = {9, 25}. That is
x2 = 9 or x2 = 25.
So x = (+/-) 3 or x = (+/-)5 or equivalent the solution set S = { -3, +3,
-5, +5}
Now complete the solution by checking that these values work.
Challenge I: Find the solution set S of the equation
( | x+3 | - 4 )(x+8) < 0.
Challenge II: Find the solution set S of the
equation | x2-17 | > 8.
Exercise: Add some diagrams to illustrate the answers.
Solving Inequalities with Absolute Value
Example I: Find the solution set S of the
equation |x| <5. Answer: S =
[-5, 5] = the interval from -5 to +5 with endpoints included.
Example II: Find the solution set S of the
equation |Q| =5. Answer: S = [-5,
5] = the interval from -5 to +5 with endpoints included.
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Using formulas forwards
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Skills for Algebra!
What
is a Variable?
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