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

##### 2 Answers

#### Answer:

#### Explanation:

We have

so it must be

Multiplying by

we get

expanding and collecting

using the quardatic formula we get

#### Answer:

#### 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-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

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

Using an AC method look for a pair of factors of

The pair

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.