How do you solve #1+x/(2-x)=(6x)/(2-x)# and check for extraneous solutions? Precalculus Solving Rational Equations Extraneous Solutions 1 Answer Mark D. Apr 10, 2018 #x=1/3# Explanation: Multiply everything by #(2-x)# to get rid of the fractions #1(2-x)+x=6x# #=># #2-x+x=6x# #=># #2=6x# #x=2/6# #x=1/3# 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 1606 views around the world You can reuse this answer Creative Commons License