How do you solve #(x-7)/(x-4)+(x+8)/(x-3)=(x+158)/(x^2-7x+12)# and check for extraneous solutions?

1 Answer
Oct 11, 2016

Answer:

#x=(7+-sqrt1401)/4#

Explanation:

As #(x-4)(x-3)=x^2-7x+12# - note that as they are in denominator #x!=3# and #x!=4#

#(x-7)/(x-4)+(x+8)/(x-3)=(x+158)/(x^2-7x+12)#

#hArr((x-7)(x-3)+(x+8)(x-4))/(x^2-7x+12)=(x+158)/(x^2-7x+12)#

or #((x-7)(x-3)+(x+8)(x-4))=(x+158)#

or #x^2-10x+21+x^2+4x-32=x+158#

or #2x^2-6x-11-x-158=0#

or #2x^2-7x-169=0#

and using quadratic formula #x=(-b+-sqrt(b^2-4ac))/(2a)# as olution for quadratic equation #ax^2+bx+c=0#

#x=(-(-7)+-sqrt((-7)^2-4*2^*(-169)))/4#

= #(7+-sqrt(49+1352))/4=(7+-sqrt1401)/4#