How do you solve #(x-6)/(x-4)+(x-7)/(x-3)=(x+111)/(x^2-7x+12)# and check for extraneous solutions?

1 Answer
Oct 1, 2016

Answer:

#x = -2 1/2 " or "x = 13#

Explanation:

In an equation with fractions, you can get rid of the fractions by multiplying every term by the LCD.

#(x-6)/(x-4)+(x-7)/(x-3)=(x+111)/(x^2-7x+12)" "larr# factor the denominator

#(x-6)/(x-4)+(x-7)/(x-3)=(x+111)/((x-4)(x-3))" "LCD = color(red)((x-4)(x-3))#

In the interests of space, let #color(red)((x-4) = Aand (x-3) = B#

#(color(red)(cancelA*B)xx(x-6))/cancel(x-4)+(color(red)(A*cancelBxx)(x-7))/cancel(x-3)=(cancelcolor(red)(A*Bxx)(x+111))/cancel((x-4)(x-3))#

#(x-3)(x-6) + (x-4)(x-7) = x+111" "larr# no fractions!

#x^2-9x +18 + x^2-11x+28 = x+111" "larr# no brackets!

#2x^2-20x-x +46 -111 = 0" "larr# make the quadratic = 0

#2x^2-21x-65 = 0" "larr# simplify and then find factors

#(2x+5)(x-13) = 0" "larr# solve with each factor =0

If #2x +5 = 0 " "rarr x = -2 1/2#
If #x-13 =0 " "rarr x = 13#