YOU are better than YOU think. Show yourself  how: 

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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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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.

Absolute Value Equations and Inequalities

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)

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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.

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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. 

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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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