how I solve this : (5x-4)(2x+3)>0 and x-3/3x+4

We'll solve the 1st
inequality:

(5x-4)(2x+3)>0

We'll
remove the brackets:

10x^2 + 15x - 8x - 12 >
0

10x^2 + 7x - 12 >
0

We'll determine the roots of the quadratic 10x^2 + 7x -
12 = 0.

x1 = [-7+sqrt(49 +
480)]/20

x1 = (-7+23)/20

x1 =
16/20 => x1 = 4/5

x2 = -30/20 => x2 =
-3/2

The expression is positive if x belongs to the reunion
of sets: (-infinite ; -3/2)U(4/5 ; +infinite).

We'll solve
the 2nd inequality:

(x-3)/(3x+4)<=2 (Please, pay
attention to the manner of writting a fraction)

We'll
subtract 2 both sides:

(x-3)/(3x+4) - 2 <=
0

(x - 3 - 6x - 8)/(3x + 4) <=
0

We'll combine like
terms:

(-5x - 11)/(3x + 4) <=
0

(5x + 11)/(3x + 4) >=
0

The fraction is positive if x belongs to the reunion of
intervals: (-infinite ; -11/5]U(-4/3 ;
+infinite)

Since the final solution has to
make both inequalities to hold, therefore x belongs to the reunion of intervals:
(-infinite ; -11/5]U(4/5 ; +infinite).