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
May 22, 2018

Answer:

#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#

Aug 14, 2018

Answer:

#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.