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1 alone is not sufficent 2 alone says they ether both positive or both negative lets continue with 2 in the case they are both negative: x<y lets take x=-4 , y= -3.5 if we put it on the (1) we get x=0.5+y --> -4 = 0.5 - 3.5 --> -4 = -3 so it can't be made. if both positive (1) is solve able hence th...
- by tzvister
Wed Mar 25, 2009 5:34 pm- Forum: Problem Solving
- Topic: anyone can help on this?
- Replies: 3
- Views: 1146
Just multiply the -x+2 part by (-1):
3x^2(x-2)+(-1)(x-2) / x-2
we get (3x^2-1)(x-2) / x-2
cancel nominator and denominator x-2
we get
3x^2-1
- by tzvister
Wed Mar 25, 2009 5:16 pm- Forum: Problem Solving
- Topic: Tricky algebra
- Replies: 3
- Views: 2070