Question

How do I solve this?

Answer 1

Case 1: $\frac{3}{7} < \frac{4 - x}{2x + 5}$

$\frac{3}{7} < \frac{(4 - x)(2x + 5)}{(2x + 5)^{2}}$
$\frac{3(2x + 5)^{2}}{7} < (4 - x)(2x + 5)$
$3(2x + 5)^{2} < 7(4 - x)(2x + 5)$
$3(4x^{2} + 20x + 25) < 7(3x + 20 - 2x^{2})$
$12x^{2} + 60x + 75 < 21x + 140 - 14x^{2}$
$26x^{2} + 39x - 65 < 0$
$13(2x^{2} + 3x - 5) < 0$
$2x^{2} + 3x - 5 < 0$
$(2x + 5)(x - 1) < 0$

Thus, we know that $-\frac{5}{2} < x < 1$ for case 1.

Case 2: $\frac{4 - x}{2x + 5} < \frac{4}{5}$
$(4 - x)(2x + 5) < \frac{4(2x + 5)^{2}}{5}$
$5(4 - x)(2x + 5) < 4(4x^{2} + 20x + 25)$
$5(3x + 20 - 2x^{2}) < 4(4x^{2} + 20x + 25)$
$15x + 100 - 10x^{2} < 16x^{2} + 80x + 100$
$0 < 26x^{2} + 65x$
$0 < 13x(2x + 5)$

Case 2: $x > 0, x < -\frac{5}{2}$

Thus, the only scenario where both cases satisfy are: $0 < x < 1$