How do you solve #\frac { x } { x + 6} - \frac { 4} { x + 9} = \frac { 1x - 12} { x ^ { 2} + 15x + 54}#?

2 Answers
Noah G
Nov 2, 2016

#x = 2#.

Explanation:

Start by factoring #x^2 + 15x + 54# to see what will be our LCD (Least Common Denominator).

#x^2 + 15x+ 54 = (x + 6)(x + 9)#

#:.#Our LCD is #(x + 6)(x + 9)#. In other words, to be equivalent, all fractions in this equation need to have a denominator of #(x + 6)(x + 9)#.

#(x(x + 9))/((x + 6)(x + 9)) - (4(x + 6))/((x + 6)(x + 9)) = (x - 12)/((x + 6)(x + 9))#

We can now eliminate the denominators.

#x^2 + 9x - 4x - 24 = x - 12#

#x^2 + 5x - 24 = x - 12#

#x^2 + 4x - 12 = 0#

#(x + 6)(x - 2) = 0#

#x = -6 and 2#

However, #x = -6# is an extraneous solution, since it renders the original equation undefined (the initial equation had restrictions on the variable of #x!= -9, -6#).

Hopefully this helps!

Barney V.
Jan 25, 2017

#x=2#

Explanation:

#x/(x+6)-4/(x+9)=(x-12)/(x^2+15x+54#

#:.x/(x+6)-4/(x+9)=(x-12)/((x+6)(x+9))#

#:.(x(x+9)-4(x+6)=x-12)/((x+6)(x+9)#

#:.(x(x+9))/((x+6)(x+9))-(4(x+6))/((x+6)(x+9))=(x-12)/((x+6)(x+9)#

#:.(x^2+9x-4x-24=x-12)/((x+6)(x+9))#

#:.(x^2+5x-24=x-12)/((x+6)(x+9))#

#:.(x^2+5x-24+12-x=0)/((x+6)(x+9))#

#:.(x^2+4x-12=0)/((x+6)(x+9))#

#:.(cancel((x+6))(x-2)=0)/(cancel(x+6)(x+9))#

#:.(x-2)/(x+9)=0#--------(1)

multiply both sides by(x+9) in (1)

#:.x-2=0#

#:.x=2#

multiply both sides by #1/(x-2)# in (1)
#:.x+9=0#

#:.x=-9# an extraneous solution

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