# how I solve this : (5x-4)(2x+3)>0 and x-3/3x+4<=2 Please I need procedure for that!

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### 1 Answer

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).**