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Topic: I need to review modulus function
Replies: 4   Last Post: Jul 23, 2013 11:19 PM

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grei

Posts: 128
Registered: 11/27/12
Re: I need to review modulus function
Posted: May 16, 2013 10:09 PM
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If you know the modulus function, then you know that |x|= x if x>= 0, |x|= -x if x< 0. So, here, the function will have different forms when x-3>= 0 and when x- 3< 0.

When x- 3>=0, in other words, when x>= 3, |x- 3|= x- 3 so the equation becomes 10- 9(x- 3)= 10- 9x+ 27= 37- 9x= 12 so 9x= 25. That is true for x= 25/9= 2 and 7/9. But that is less than 3 and so is not a valid solution.

When x- 3< 0, in other words, when x< 3, |x- 3|= -(x- 3)= -x+ 3 so the equation becomes 10- 9(-x+ 3)= 10+ 9x- 27= 9x- 17= 12 sp 9x= 29. That is true for x= 29/9= 3 and 2/9. But that is larger than 3 and so is not a valid solution.

There is NO value of x satisfying 10- 9|x- 3|= 12.



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