+817 I cant solve this it's hard
Solve the inequality
\[\dfrac{x+1}{x+2}>\dfrac{3x+4}{2x+9}+\dfrac{1}{3}.\]
+129847 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 !!!
