Question

For what `x` is `3/(x+3)>(4x)/(x-2)`?

Answer 1

`x\in(-3, 2)`

Explanation:

To simplify the inequality, we are going to perform cross-multiplication of the denominator to avoid dealing with fractions. However, the sign of the inequality might be flipped if we multiply both sides by a negative expression. Hence, we consider `x` on 3 separate intervals, namely, `(-oo,-3)`, `(-3, 2)`, `(2, oo)`. Note that `x` cannot -3 or 2. Now, we wish to find the values of `x` which satisfies

`frac{3}{x+3}>frac{4x}{x-2}`

For the case of `x\in(-oo, -3)`, both `x+3` and `x-2` are negative.

`3(x-2)>4x(x+3)`

The sign remains the same as the product of 2 negative expressions is positive.

`3x-6>4x^2+12x`

`0>4x^2+9x+6`

`0>x^2+9/4x+3/2`

Completing the square,
`0>(x+9/8)^2+{15}/{64}`

Since the RHS is always `>=15/64`, no real value of `x` satisfies
`3(x-2)>4x(x+3)`.

Therefore, no value of `x\in(-oo, -3)` satisfies `frac{3}{x+3}>frac{4x}{x-2}`.

For the case of `x\in(-3, 2)`, `x+3` is positive while `x-2` is negative.

`3(x-2)<4x(x+3)`

The sign flips as the product of a positive and a negative expressions is negative.

Completing the square,
`0<(x+9/8)^2+{15}/{64}`

Since the RHS is always `>=15/64>0`, all real value of `x` satisfies
`3(x-2)>4x(x+3)`.

Therefore, all values of `x\in(-3, 2)` satisfies `frac{3}{x+3}>frac{4x}{x-2}`.

For the case of `x\in(2, oo)`, both `x+3` and `x-2` are negative.

`3(x-2)>4x(x+3)`

The sign remains the same as the product of 2 negative expressions is positive.

Completing the square,
`0>(x+9/8)^2+{15}/{64}`

Since the RHS is always `>=15/64`, no real value of `x` satisfies
`3(x-2)>4x(x+3)`.

Therefore, no value of `x\in(2, oo)` satisfies `frac{3}{x+3}>frac{4x}{x-2}`.

Combining the solutions on all 3 intervals, we get

`frac{3}{x+3}>frac{4x}{x-2}`

if and only if `x\in(-3, 2)`.