Question

How do you solve `\frac { 9} { x + 3} - \frac { 5} { x + 4} = \frac { - 11} { x ^ { 2} + 7x + 12}`?

Answer 1

`color(red)(x=-8)`

Explanation:

If we note that `(x+3)xx(x+4)=(x^2+7x+12)`

we can convert the given equation:
`color(white)("XXX")color(blue)(9/(x+3))-color(magenta)(5/(x+4))=color(green)((-11)/(x^2+7x+12))`

into a form where all the terms have the same denominator:
`color(white)("XXX")color(blue)(9/(x+3)xx(x+4)/(x+4))-color(magenta)(5/(x+4)xx(x+3)/(x+3))=color(green)((-11)/(x^2+7x+12))`

`color(white)("XXX")color(blue)((9x+36)/(x^2+7x+12))-color(magenta)((5x+15)/(x^2+7x+12))=color(green)((-11)/(x^2+7x+12))`

If we make the (at least temporary) assumption that `x^2+7x+12!=0`
then we can multiply everything by `x^2+7x+12`
to get
`color(white)("XXX")color(blue)(""(9x+36))-color(magenta)(""(5x+15))=color(green)(-11)`

Simplifying:
`color(white)("XXX")4x+21=color(green)(-11)`

`color(white)("XXX")4x=-32`

`color(white)("XXX")x=-8`

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There are a couple things we should check at this point:

1. that our assumption that `x^2+7x+12!=0` for this value of `x` (in fact it is equal to `20` as we can see with a quick substitution);
2. that `x=-8` when substituted into our original equation does give a valid equation (this is basically a check that we haven't made any errors in our calculations; again doing the substitution we get `-5/9` on both sides of the equation, so our result is valid).