Question

How do you solve `6x - 5 < 6/x`?

Answer 1

`x in (-2/3, 0) uu (0, 3/2)`

Explanation:

Right from the start, you know that `x` cannot be equal to zero, since it is the denominator of a fraction. So you need to have `x!=0`.

With that in mind, multiply the left-hand side of the inequality by `1= x/x` to get rid of the denominator

`6x * x/x - 5 * x/x < 6x`

`6x^2 - 5x < 6`

Next, add `-6` to both sides of the inequality

`6x^2 - 5x - 6 < color(red)(cancel(color(black)(6))) - color(red)(cancel(color(black)(6)))`

`6x^2 - 5x - 6 < 0`

To help you determine the intervals on which this quadratic function is smaller than zero, you need to first determine its root by using the quadratic formula

`6x^2 - 5x - 6 = 0`

`x_(1,2) = (-(-5) +- sqrt((-5)^2 - 4 * 6 * (-6)))/(2 * 6)`

`x_(1,2) = (5 +- sqrt(169))/12`

`x_(1,2) = (5 +- 13)/12 = {(x_1 = (5 + 13)/12 = 3/2), (x_2 = (5 - 13)/12 = -2/3) :}`

You can thus rewrite the quadratic as

`6(x-3/2)(x+2/3) = 0`

So, you need this expression to be negative, which implies that `(x-3/2)` must be positive whenever `(x+2/3)` is negative, and vice versa.

For `x> -2/3` and `x<3/2` you get

`{(x-3/2 < 0), (x + 2/3 > 0) :} implies (x-3/2)(x+2/3) < 0`

Any value of `x>3/2` will make both terms positive, and any value of `x<-2/3` will make both terms negative. Keeping in mind that you also need `x!=0`, the solution set for this inequality will be `x in (-2/3, 0) uu (0, 3/2)`.