Question

Va rog frumos am nevoie rapid

Answer 1

a)
E1 < E2 <=> (2x-5)/3 < -(4x+1)/5 <=> 10x-25 < -12x-3 <=> 22x < 22 <=> x < 1

b)
E1+E2 = (2x-5)/3-(4x+1)/5 = [(10x-25)-(12x+3)]/15 = -(2x+28)/15

E1+E2 >= 0 <=> -(2x+8)/15 >= 0 <=> 2x+8 <= 0 <=> x <= -4.

c)
E1 > 0 <=> (2x-5)/3 > 0 <=> 2x-5 > 0 <=> x > 5/2.
E2 <= 0 <=> -(4x+1)/5 <= 0 <=> 4x+1 >= 0 <=> x >= -1/4.

Deci E1 > 0 si E2 <= 0 daca si numai daca x > 5/2.