1. ## Inequality :\

Stupid inequality i can't solve, we've been told to show a certain method when solving the question which i don't understand T_T

Question
$\frac{3}{3x-2}>\frac{1}{x+4}$

The method we are meant to use is supposed to involved these if statements (4 of them)

1. If (x+4)(3x-2) are +ve then ans is ....
2. If (x+4)(3x-2) are -ve then ans is ....
3. If (x+4) is +ve and (3x-2) is -ve then ans is ....
4. If (3x-2) is +ve and (x+4) is -ve then ans is ....

I think i can do the 1st if statement
$\frac{3}{3x-2}>\frac{1}{x+4}$
If both denominators +ve multiply both sides by (x+4)(3x-2) which results in.
$3(x+4)>3x-2$
$3x+12>3x-2$
$14>0$

Got no idea how to do 2.3.4 someone help meee

1.

2. Originally Posted by Kevlar

1. If (x+4)(3x-2) are +ve then ans is ....
The $>$ will not change

Originally Posted by Kevlar

2. If (x+4)(3x-2) are -ve then ans is ....
If both terms are negative then the product will be positive and
the $>$ will not change.

If $(x+4)(3x-2)$ is all negative then the $>$ will change to $<$

Originally Posted by Kevlar

3. If (x+4) is +ve and (3x-2) is -ve then ans is ....
The $>$ will change to $<$ as the product is negative

Originally Posted by Kevlar

4. If (3x-2) is +ve and (x+4) is -ve then ans is ....
The $>$ will change to $<$ as the product is negative

3. Ok ima try the 2nd and 3rd ones then

2. If (x+4) and (3x-2) are -ve then

$\frac{3}{3x-2}>\frac{1}{x+4}$
is multiplied on both sides by -(3x-2).-(x+4)

$-3.-(x+4)>-1.-(3x-2)$
$3x+13>3x-2$
$14>0$

???? :z graphs dont intersect if both are +ve or both are -ve? :z

3. If (x+4) is +ve, (3x-2) is -ve.

$\frac{3}{3x-2}>\frac{1}{x+4}$
is multiplied on both sides by -(3x-2).(x+4)

$-3.(x+4)<-(3x-2)$
$-3x-12<-3x+2$
$-14<0$

i think ive gone wrong arghttt