How do you solve #10/(n^2-1)+(2n-5)/(n-1)=(2n+5)/(n+1)#? Precalculus Solving Rational Equations Extraneous Solutions 1 Answer Noah G Feb 22, 2017 #n = 5/3# Explanation: Note that #n^2 - 1 = (n + 1)(n - 1)#. #10/((n + 1)(n - 1)) + ((2n - 5)(n + 1))/((n + 1)(n - 1)) = ((2n + 5)(n - 1))/((n - 1)(n + 1))# #10 + 2n^2 - 5n + 2n - 5 = 2n^2 + 5n - 2n - 5# #10 = 6n# #5/3 = n# Hopefully this helps! Answer link Related questions What are extraneous solutions? What are common mistakes students make with respect to extraneous solutions? How do extraneous solutions arise? How do extraneous solutions arise from radical equations? How do I check for extraneous solutions? What are some examples of extraneous solutions to equations? How do I find the extraneous solution of #sqrt(x+4)=x-2#? How do I find the extraneous solution of #y-5=4sqrt(y)#? How do I find the extraneous solution of #sqrt(x-1)=x-7#? How do I find the extraneous solution of #sqrt(x-3)-sqrt(x)=3#? See all questions in Extraneous Solutions Impact of this question 2419 views around the world You can reuse this answer Creative Commons License