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grei
Posts:
132
Registered:
11/27/12


Re: I need to review modulus function
Posted:
May 16, 2013 10:09 PM


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 x3>= 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 9x 3= 12.



