how do u solve ( x + 1) / (x - 2) <2 ? please help?

(x+1)/(x-2) - 2 < 0
(x + 1 - 2x + 4) / (x-2) < 0
(5 - x)/(x-2) < 0

(x-5)/(x-2) > 0

thus this will be valid if the two expressions are of the same sign...

the possible answers are
x < 2 or x > 5.

§

x + 1 < 2x - 4
5 < x
x > 5

|x - a million| + |x + 2| < 5 sq. the two facets, (x - a million)^2 + 2|x - a million||x + 2| + (x + 2)^2 < 5^2 2x^2 + 2x + 5 + 2|x - a million||x + 2| < 25 2x^2 + 2x + 2|x - a million||x + 2| < 20 x^2 + x + |x - a million||x + 2| < 10 (x + 0.5)^2 + |x - a million||x + 2| < 10.25 |x - a million||x + 2| = {[(x - a million) + (x + 2)]/2}^2 - {[(x - a million) - (x + 2)]/2}^2 = [x + 0.5]^2 - a million.5^2 for that reason, (x + 0.5)^2 + (x + 0.5)^2 - 2.25 < 10.25 2(x + 0.5)^2 < 12.5 x^2 + x + 0.25 < 6.25 x^2 + x - 6 < 0 (x + 3)(x - 2) < 0 -3 < x < 2

(x + 1)/(x - 2) < 2
(x - 2)[(x + 1)/(x - 2)] < 2(x - 2)
x + 1 < 2*x - 2*2
x + 1 < 2x - 4
x - 2x < -1 - 4
-x < -5
x < -5/-1
x < 5

I cannot multiply by (x-2( as it could be positive or -ve . so I multiply by (x-2)^2
(x+1)(x-2) < 2 (x-2)^2
take the 2(x-2) to the left
(x-2)(x+1-2x +4) < 0
(x-2)(5-x) <0
(x-2)(x-5) >0
both are positive or -ve
x >5 or < 2

x>5

(x + 1) / (x + 2) < 2
x + 1 < 2x - 4
x - 2x < -5
- x < -5
x > 5


;) I hope that explains it all. should be easy enough to understand.

Since there is an x in the denominator, we can't just multiply across, because (x-2) might be negative which will flip the sign of the inequality. What we do instead, is to multiply by the square of the denominator which must be positive
(x+1)/(x-2) < 2
(x+1)(x-2) < 2(x-2)^2
(x+1)(x-2) - 2(x-2)^2 < 0
(x-2)[(x+1) - 2(x-2) < 0
(x-2)(x+1-2x+4) < 0
- (x-2)(x-5) < 0
(x-2)(x-5) > 0
if we graph this quadratic whose roots are at x=2 and x = 5
...++++2--------5++++++...
so the ineq holds for x < 2, x > 5

greater than 5

The technique is never cross multiply.
Thus, you have to transpose 2 from the right hand side of the inequality to the lefthand side.
((x+1)/(x-2))-2<0.

Then simplify the lefthand side: (x+1-2x+4)/(x-2)<0
(-x+5)/(x-2)<0. Then we have to determine the domain of x for which the last inequality would be true. This is true if the numerator is positive and the denominator is negative or if the numerator is negative and the denominator is positive.
Case 1: Numerator is positive and denominator is negative.
-x+5>0 and x-2<0. Since -x+5>0 implies 5>x and x-2<0 implies x<2. We find the intersection of these two inequalities(i.e. which values of x satisfies both inequalities). Any value of x less than 2 would satisfy both inequality. Thus, the answer for Case 1 is x<2.

Case 2: Numerator is negative and denominator is positive.
-x+5<0 and x-2>0. These imply x>5 and x>2. Again we find the intersection which is x>5.

Therefore the answer for this would be, all values greater than 5 and all values less than 2 (x>5 U x<2).