Thread: Help Inequalities with absolute

8.) /(3-2x) divided by (2+x)/ < 4

/ means absolute value

Originally Posted by ^_^Engineer_Adam^_^

8.) /(3-2x) divided by (2+x)/ < 4

/ means absolute value

Hi,

I haven't much time, therefore I attach a diagram so you can get graphically a solution. (I only guess: x < (-11)/2 or x > (-5)/6 )

If I find some time today the algebraic solution will follow.

tschüss

EB

Originally Posted by ^_^Engineer_Adam^_^

8.) /(3-2x) divided by (2+x)/ < 4

/ means absolute value

|(3-2x)/(2+x)|<4,

means:

-4 < (3-2x)/(2+x) < 4

Now its a good idea to sketch (3-2x)/(2+x), (see attachment)

Now solving (3-2x)/(2+x) = -4, gives x=-11/2 and so

-4 < (3-2x)/(2+x)

when x<-11/2, and alsowhen x>-2 (where the vertical asymtote is).

Now solve (3-2x)/(2+x) = 4, gives x=-5/6 and so:

(3-2x)/(2+x) < 4 for x>-5/6m and also when x<-2.

Combining these we see that both inequalities are satisfied when

x<-11/2 or x>-5/6.

RonL

Originally Posted by ^_^Engineer_Adam^_^

8.) /(3-2x) divided by (2+x)/ < 4

/ means absolute value

Hi,

here I am again. You already know the solution of your problem, because CaptBlack has done it. I promised to come back with a solution. So here are my two cents:

abs((3-2x)/(2+x)) < 4 , x ≠ -2

First remove the absolute value:

A) (3-2x)/(2+x) > -4 and B) (3-2x)/(2+x) < 4

A) Multiply by the denominator. There are 2 possibilities:

A1) 3-2x > -4*(2+x) and x > -2
3 - 2x > -8 - 4x
11 > -2x
x < (-11)/2 and x > -2 that means: no solution

A2) 3-2x < -4*(2+x) and x < - 2
3 - 2x < -8 - 4x
11 < -2x
x < (-11)/2 and x < -2 that means: this set belongs to the solution


B) Multiply by the denominator. There are 2 possibilities:

B1) 3-2x < 4*(2+x) and x > -2
3 - 2x < 8 + 4x
-5 < 6x
x > (-5)/6 and x > -2 that means: this set belongs to the solution

B2) 3-2x < 4*(2+x) and x < -2
3 - 2x < 8 + 4x
-5 < 6x
x > (-5)/6 and x < -2 that means: no solution

tschüss

EB