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# solving inequalities involving the modulus sign watch

1. f) |x+3| = |1-x|
j) 3|2x-1| = |x|

how would i solve these? there isn't any examples in the book regarding them.
2. If you square both sides of each equation then you can omit the modulus signs (as (x+3)^2 and (1-x)^2 will both be positive, regardless of whether x is a positive or negative number.) Expand the brackets and rearrange all the terms onto one side, and you should be able to come up with a solution.
3. Just write it out in full, what is the definition of the modulus?

4. i also need help with a question like this... the question ive been asked is..

find all x ∈ R that satisfy the inequality:

|x-1| > 1/2
x+1

does anybody know how to solve this so i can find all possible x values????
5. (Original post by sarah_vickers)
i also need help with a question like this... the question ive been asked is..

find all x ∈ R that satisfy the inequality:

|x-1| > 1/2
x+1

does anybody know how to solve this so i can find all possible x values????
If you square both sides of the inequality you do two things: Firstly, you can get rid of the modulus sign around the (x-1) (because if x is a real number, then (x-1)^2 will be positive regardless of whether x is negative or positive). Secondly, you can multiply both sides of the inequality by (x+1)^2, leaving you with (x-1)^2 > 0.25(x+1)^2.If you mutiply both sides by four and expand the brackets, you end up with (4x^2)-8x+4 > (x^2)+2x+1. Collecting the terms on one side you get (3x^2)-10x+3>0, which factorises to (3x-1)(x-3)>0. If you sketch this graph, i.e. a positive parabola with x intersects at 1/3 and 3, you should be able to answer the question,. i.e. x<1/3 and x>3. I'll double check this later (as I've not got a graphical calculator with me at the moment and my brain is frazzled so I've probably made a mistake )
6. I do modulus in a different way.

1) draw graph of modulus

2) look for intersection of the lines

3)if line is the 'original line' ie not reflected in y axis remove modulus sign

4)If line has been reflected in y axis multiply equation for the line by -1

5 then solve like a normal inequality. ie set them equal to each other > simplify
7. (Original post by DaGianni)
If you square both sides of the inequality you do two things: Firstly, you can get rid of the modulus sign around the (x-1) (because if x is a real number, then (x-1)^2 will be positive regardless of whether x is negative or positive). Secondly, you can multiply both sides of the inequality by (x+1)^2, leaving you with (x-1)^2 > 0.25(x+1)^2.If you mutiply both sides by four and expand the brackets, you end up with (4x^2)-8x+4 > (x^2)+2x+1. Collecting the terms on one side you get (3x^2)-10x+3>0, which factorises to (3x-1)(x-3)>0. If you sketch this graph, i.e. a positive parabola with x intersects at 1/3 and 3, you should be able to answer the question,. i.e. x<1/3 and x>3. I'll double check this later (as I've not got a graphical calculator with me at the moment and my brain is frazzled so I've probably made a mistake )
yeah ive just tried this method a second before ure comment lol... i multiplied both sides by x+1 first and then squared. but for my answers >1/3 and >3

x values bigger than 3 satisfy the equation, but x values for <1/3 only satisty it up to 0, meaning 0 < x < 1/3 and x > 3 as my final answers?
8. (Original post by sarah_vickers)
yeah ive just tried this method a second before ure comment lol... i multiplied both sides by x+1 first and then squared. but for my answers >1/3 and >3

x values bigger than 3 satisfy the equation, but x values for <1/3 only satisty it up to 0, meaning 0 < x < 1/3 and x > 3 as my final answers?
Are you sure you don't mean x values down to -1 satisfy it? There is a vertical asymtote on |x-1|/(x+1) at x = -1, so as you go down from x=1/3, the value of y should increase until at x=-1 you hit the asymptote. For x values lower than -1, y will always be negative so that won't satisfy the inequality. This (http://www.wolframalpha.com/input/?i=%28|x-1|%2F%28x%2B1%29%29+%3E+0.5) gives you a diagram of the regions that satisfy your inequality.

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Updated: January 19, 2010
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