how do I solve 1+x/1-x > 2-x/2+x ?? please explain me how to solve more of these kinds.?

Question

the procedures to solve these kind of inequalities are confusing me! I would love to have a helping hand....

Thanks for answering,,, Cheers!!

Answer 1

take all terms over to one side:
(1+x)/(1-x) - (2-x)/(2+x) > 0

find a common denominator:
[ (1+x)(2+x) - (2-x)(1-x) ] / [ (1-x)(2+x) ] > 0

by rearranging the top line, we get:
6x / [ (1-x)(2+x) ] > 0

from this we can see the critical values are -2, 0, 1.
draw up a table (5 by 5):
[space] (x < -2) (-2 < x < 0) (0 < x < 1) (1 < x)
6x -ve -ve +ve +ve
1-x +ve +ve +ve -ve
2+x -ve +ve +ve +ve
overall +ve -ve +ve -ve

[table headings are determined by critical values.
to find whether 6x is +ve or -ve for x < -2, substitute in a possible value of x eg. -3 --> 6(-3) = -18 therefore -ve.
to find whether it is overall +ve or -ve, add the signs in that column, eg. for x < -2, a -ve plus a +ve = -ve, add on another -ve and it's +ve.]

since the identity asks for > 0, we want the positive results ie.
x < -2 and 0 < x < 1.

hope that helps :)
and hopefully the table makes sense...

Answer 2

Move all the Xs to the left hand side and all the numbers to the right (changing sign as you do) then simplify. This one solves in 5 lines, but it's difficult to write it out in a text box.

Answer 3

Multiply both sides by 1-x andd 2+x: This gets rid of the denominator in both fractions.

(1+x)(2+x) > (2-x)(1-x)


Now expand both sides:

x^2+3x+2>x^2-3x+2

Bring everything over to one side, and see that x^2 and 2 both cancel out:

6x>0
x>0

Answer 4

1+x/1-x > 2-x/2+x
1+x/1-x - 2-x/2+x > 0
[(1+x)(2+x) - (2-x)(1-x)] / (1-x)(2+x) > 0
[ 2 + 3x + x^2 - 2 + 3x - x^2] / (1-x)(2+x) > 0
6x / (1-x)(2+x) > 0

By drawing a number line with -2, 0 and 1,
-21

Answer 5

cross multiplying
(1+x)(2+x)>(1-x)(2-x)
x^2+3x+2>x^2-3x+2
6x>0
x>0