How do you solve #7/(x+5)+3/(x-5)=30/(x^2-25)# and check for extraneous solutions?

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Answer 1 · 2017-02-17T11:58:19

I got that it has no real solutions and only one extraneous.

Have a look:
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Answer 2 · 2017-02-17T12:07:54

Note that #x^2-25=(x+5)(x-5)#

First we assert that #x!=-5andx!=+5#

After that, we we multiply the first two terms by disguised #1#'s:

#7/(x+5)xx(x-5)/(x-5)=(7(x-5))/(x^2-25)# and:

#3/(x-5)xx(x+5)/(x+5)=(3(x+5))/(x^2-25)#

Together this adds up to:

#(7(x-5))/(x^2-25)+(3(x+5))/(x^2-25)=30/(x^2-25)->#

#(7x-35+3x+15)/(x^2-25)=30/(x^2-25)->#

#10x-20=30->10x=50->x=+5#

And this is contrary to our first assertion.

Conclusion: there is no solution.