Question

How do I solve this?

Answer 1

Case 1:
`(3)/(7) < (4 - x)/(2x + 5)`


`(3)/(7) < ((4 - x)(2x + 5))/((2x + 5)^(2))`

`(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
`-(5)/(2) < x < 1` for case 1.

Case 2:
`(4 - x)/(2x + 5) < (4)/(5)`

`(4 - x)(2x + 5) < (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 < -(5)/(2)`

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