+0  
 
-2
40
2
avatar+819 

I cant solve this it's hard

 

Solve the inequality
\[\dfrac{x+1}{x+2}>\dfrac{3x+4}{2x+9}+\dfrac{1}{3}.\]

 Aug 30, 2023
 #1
avatar+128732 
+1

Simplifying the RIGHT side we have

 

(3x + 4)  /( 2x + 9)  + 1/3  =    [ 3 (3x + 4) + 2x + 9 ]  / [ 6x + 27]  

 

[ 9x + 12 + 2x + 9 ] / [ 6x + 27]  =  [ 11x + 21] / [ 6x + 27]  ...so.....

 

[x + 1] / [x + 2] > [11x + 21]  [ 6x + 27]

 

[x + 1] / [ x + 2 ] - [ 11x + 21] [ 6x + 27 ]  >  0

 

 ( [ x + 1] [ 6x + 27]  - [ 11x + 21 ] [ x + 2] )  /  ( [ x + 2] [ 6x + 27 ])  > 0

 

[ 6x^2 + 33x + 27 - 11x^2 - 43x - 42 ] / ([ x+2] [ 6x + 27])   >  0

 

[ -5x^2 - 10x - 15 ] / [  [x + 2] * 3 * [ 2x + 9 ]  >  0

 

[ -5 (x^2 + 2x + 3) ] / [ 3 * (x + 2) * ( 2x + 9) ]  >  0     divide through by -3/5...reverse the inequality sign

 

(x ^2 + 2x + 3)  / [ (x + 2) ( 2x + 9) ] <  0

 

The top can  never be  < 0  for any  x

 

Therefore.....to make this  < 0  we need to  find the interval  where the denominator < 0

 

(x + 2) ( 2x + 9)  <   0

 

Make this an equality  ......the answer  will lie  between  the  roots

 

The roots are   -9/2  and  -2

 

So.....the answer is   (-9/2 ,  -2)

 

CORRECTED  !!!

 

cool cool cool

 Aug 30, 2023
edited by CPhill  Aug 30, 2023
edited by CPhill  Aug 30, 2023
edited by CPhill  Aug 30, 2023

4 Online Users

avatar