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3 Answers
- icemanLv 78 years agoFavourite answer
(x-2) / (x+2) > (2x-3) / (4x-1)
(x-2) / (x+2) - (2x-3) / (4x-1) > 0
[(x - 2)(4x - 1) - (2x - 3)(x + 2)] /[(x + 2)(4x - 1)] > 0
(2x^2 - 10x + 8)/[(x + 2)(4x - 1)] > 0
[2(x - 1)(x - 4)]/[(x + 2)(4x - 1)] > 0
analysis: for this fraction to be positive the numerator and the denominator must be both positive or both negative:
1) both positive:
(x - 1)(x - 4) > 0
x < 1 or x > 4
[(x + 2)(4x - 1) > 0
x < -2 or x > 1/4
2) both negative:
(x - 1)(x - 4) < 0
1 < x < 4
(x + 2)(4x - 1) < 0
-2 < x < 1/4
combine both case we get:
x > 4 or 1/4 < x < 1 or x < -2
in interval form:
(-∞ , -2) U (1/4 , 1) U (4 , ∞)
I hope this helps.
- 8 years ago
x-2/x+2 > 2x-3/4x-1
= (x-2)(4x-1) > (x+2)(2x-3)(by cross-multiplication)
= 4x^2 - 9x + 2 > 2x^2 + x -6
= 2x^2 - 10x + 8 >0
dividing by 2 on both sides,
x^2 - 5x + 4 >0
= x^2 - x - 4x +4 >0
=x(x-1) - 4(x-1) >0
=(x-4)(x-1)>0,x belongs to (-infinity,1) U (4,infinity),x iz not equal to -2,1/4