Question

How do you solve `(x-4)/(x-7)+(x-5)/(x-6)=(x+164)/(x^2-13x+42)` and check for extraneous solutions?

Answer 1

`x_1=15/2` or `x_2=-7/2`

Explanation:

We have
`(x-7)(x-6)=x^2-13x+42`
so it must be
`xne 7,x\ne6`
Multiplying by`(x-6)(x-7)`
we get
`(x-4)(x-6)+(x-5)(x-7)=x+164`
expanding and collecting
`2x^2-22x+59=x+164`
`2x^2-22x-105=0`
`x^2-23/2x-105/2=0`
using the quardatic formula we get
`x_1=15`
`x_2=-7/2`

Answer 2

`x=-7/2" "` or `" "x=15`

Explanation:

Given:

`(x-4)/(x-7)+(x-5)/(x-6)=(x+164)/(x^2-13x+42)`

Note that:

`x^2-13x+42 = (x-7)(x-6)`

So multiplying both sides of the given equation by `(x-7)(x-6)` we get:

`(x-4)(x-6)+(x-5)(x-7) = x+164`

Multiplied out that becomes:

`x^2-10x+24+x^2-12x+35 = x+164`

That is:

`2x^2-22x+59 = x+164`

Subtracting `x+164` from both sides, that becomes:

`2x^2-23x-105 = 0`

Using an AC method look for a pair of factors of `AC = 2 * 105 = 210` which differ by `B=23`.

The pair `30, 7` works.

Use this pair to split the middle term and factor by grouping:

`0 = 2x^2-23x-105`

`color(white)(0) = (2x^2-30x)+(7x-105)`

`color(white)(0) = 2x(x-15)+7(x-15)`

`color(white)(0) = (2x+7)(x-15)`

Hence:

`x=-7/2" "` or `" "x=15`

Note that neither of these values cause a zero denominator in the given rational equation, so they are both valid solutions.