How do you solve `(x+7)/(x-4)<0`?

For `(x+7)/(x-4)` to be less than `0`, the top part must be less than `0`, or negative.

An `x` value of `-7` will cancel the top part, but only give us `0`, anything less than `-7` will give a negative fraction, and therefore less than `0`.

Proof:
`(x+7)/(x-4)lt0`

`((x+7)cancel((x-4)))/cancel((x-4))lt0(x-4)`

`x+7lt0`

`x+7-7lt-7`

`xlt-7`

However, if top and bottom are negative, it will be positive, and any `x` value less than `-7` will give a positive value. However if `xlt4`, then the bottom value will be negative, giving a negative value.

`-7ltxlt4`

Let `f(x)=(x+7)/(x-4)`

We can build the sign chart

`color(white)(aaaa)``x``color(white)(aaaa)``-oo``color(white)(aaaa)``-7``color(white)(aaaaaa)``4``color(white)(aaaaaa)``+oo`

`color(white)(aaaa)``x+7``color(white)(aaaa)``-``color(white)(aaa)``0``color(white)(aa)``+``color(white)(aaaaa)``+`

`color(white)(aaaa)``x-4``color(white)(aaaa)``-``color(white)(aaaaaa)``-``color(white)(aa)``||``color(white)(aa)``+`

`color(white)(aaaa)``f(x)``color(white)(aaaaa)``+``color(white)(aaa)``0``color(white)(aa)``-``color(white)(aa)``||``color(white)(aa)``+`

Therefore,

`f(x)<0` when `x in (-7,4)`

graph{(x+7)/(x-4) [-41.1, 41.14, -20.54, 20.55]}

Given:

`(x+7)/(x-4) < 0`

Note that since the linear expressions `(x+7)` and `(x-4)` each occur once, the rational expression will change sign at the points `x=-7` and `x=4`. It has a vertical asymptote at `x=4` and intercepts the `x` axis at `x=-7`.

For large positive or negative values of `x`, the rational expression is positive, so the interval in which it is negative is precisely `(-7, 4)`

graph{(y-(x+7)/(x-4))(x-3.99+y*0.0001) = 0 [-19.55, 20.45, -10.12, 9.88]}

First we have to calculate the domain of the rational expression. As the denominator cannot be zero, the excluded values are:

`x-4 != 0 => x!=4`

Now we can solve the inequality.

`(x+7)/(x-4)<0`

We can change the rational inequality to quadratic inequality by multiplying it by the square of the denominator:

`(x+7)(x-4)<0`

If we graph the quadratic function:

graph{(x-4)*(x+7) [-36.52, 36.52, -18.22, 18.35]}

we see that it takes negative values for `x in (-7;4)`, so this interval is also the solution of the initial rational inequality