How do you solve #x+(x+5)/(x-7)=(10x-58)/(x-7)# and check for extraneous solutions?

1 Answer
May 20, 2017

Answer:

9; read below for extraneous

Explanation:

First of all, seeing how this question deals with fractions, multiply the value of #x# by #(x-7)/(x-7)# to get #(x^2-7x)/(x-7)#.

Now we have three fractions with the same denominator:

#(x^2-7x)/(x-7) + (x+5)/(x-7) = (10x-58)/(x-7)#

Because the denominators are all the same, they eventually cancel out by cross-multiplication, so we can drop them to get a simple equation:

#(x^2-7x)+ (x+5) = 10x-58#

At this point, it is a matter of combining like terms and simplifying for an answer

#(x^2-7x)+ (x+5) = 10x-58#
=> #x^2 -6x +5 = 10x-58#
=>#x^2 -16x + 63 = 0#
=> #(x-9)(x-7) = 0#
#x = 7,9#

Now we have two answers, and the problem seems over, right? Well, it isn't, as we have an extraneous solution: 7. This is because, when plugged back into the original equation, 7 creates a denominator of 0; dividing by zero is impossible, so we exclude 7 as a possible answer. So we are left with one final answer: 9