Case 1: [tex]\frac{3}{7} < \frac{4 - x}{2x + 5}[/tex]
[tex]\frac{3}{7} < \frac{(4 - x)(2x + 5)}{(2x + 5)^{2}}[/tex]
[tex]\frac{3(2x + 5)^{2}}{7} < (4 - x)(2x + 5)[/tex]
[tex]3(2x + 5)^{2} < 7(4 - x)(2x + 5)[/tex]
[tex]3(4x^{2} + 20x + 25) < 7(3x + 20 - 2x^{2})[/tex]
[tex]12x^{2} + 60x + 75 < 21x + 140 - 14x^{2}[/tex]
[tex]26x^{2} + 39x - 65 < 0[/tex]
[tex]13(2x^{2} + 3x - 5) < 0[/tex]
[tex]2x^{2} + 3x - 5 < 0[/tex]
[tex](2x + 5)(x - 1) < 0[/tex]
Thus, we know that [tex]-\frac{5}{2} < x < 1[/tex] for case 1.
Case 2: [tex]\frac{4 - x}{2x + 5} < \frac{4}{5}[/tex]
[tex](4 - x)(2x + 5) < \frac{4(2x + 5)^{2}}{5}[/tex]
[tex]5(4 - x)(2x + 5) < 4(4x^{2} + 20x + 25)[/tex]
[tex]5(3x + 20 - 2x^{2}) < 4(4x^{2} + 20x + 25)[/tex]
[tex]15x + 100 - 10x^{2} < 16x^{2} + 80x + 100[/tex]
[tex]0 < 26x^{2} + 65x[/tex]
[tex]0 < 13x(2x + 5)[/tex]
Case 2: [tex]x > 0, x < -\frac{5}{2}[/tex]
Thus, the only scenario where both cases satisfy are: [tex]0 < x < 1[/tex]
